1
00:00:01,143 --> 00:00:02,874
- [Voiceover] Let's
talk about specific heat

2
00:00:02,874 --> 00:00:05,109
and the heat of fusion and vaporization.

3
00:00:05,109 --> 00:00:07,041
Let's say you had a
container of some liquid

4
00:00:07,041 --> 00:00:08,802
and you wanted to
increase the temperature,

5
00:00:08,802 --> 00:00:10,345
you'd probably add heat.

6
00:00:10,345 --> 00:00:12,135
But how much heat should you add?

7
00:00:12,135 --> 00:00:13,367
There's a formula for it.

8
00:00:13,367 --> 00:00:15,098
Let's try to figure out
what should go in here.

9
00:00:15,098 --> 00:00:16,878
It's gonna depend on a few things.

10
00:00:16,878 --> 00:00:18,547
For one it's gonna depend on

11
00:00:18,547 --> 00:00:20,902
how much you want to
increase the temperature.

12
00:00:20,902 --> 00:00:23,510
So, what is the amount
by which you want to

13
00:00:23,510 --> 00:00:24,836
increase the temperature?

14
00:00:24,836 --> 00:00:26,674
The more you wanna
increase the temperature

15
00:00:26,674 --> 00:00:28,141
the more heat you'll need to add.

16
00:00:28,141 --> 00:00:31,542
It's also gonna depend on how
much of the material you have.

17
00:00:31,542 --> 00:00:33,946
In other words, the mass
of this liquid in here.

18
00:00:33,946 --> 00:00:36,776
The more you have the more
heat you're gonna have to add

19
00:00:36,776 --> 00:00:38,078
to change that temperature.

20
00:00:38,078 --> 00:00:39,669
And it depends on one more thing,

21
00:00:39,669 --> 00:00:42,472
the specific heat of
that particular material.

22
00:00:42,472 --> 00:00:44,579
So, different materials will be harder to

23
00:00:44,579 --> 00:00:47,042
increase in temperature
than other materials.

24
00:00:47,042 --> 00:00:49,500
And if a material has a high specific heat

25
00:00:49,500 --> 00:00:51,466
it'll take more joules of heat

26
00:00:51,466 --> 00:00:53,337
in order to increase the temperature.

27
00:00:53,337 --> 00:00:55,041
So let's get a little more specific.

28
00:00:55,041 --> 00:00:56,601
Let's say our liquid was water

29
00:00:56,601 --> 00:01:00,375
and it was at a temperature
of 20 degrees Celsius.

30
00:01:00,375 --> 00:01:01,870
Let's say it was a big container that had

31
00:01:01,870 --> 00:01:03,913
two kilograms of water in it.

32
00:01:03,913 --> 00:01:05,146
Now, the specific heat,

33
00:01:05,146 --> 00:01:07,079
you can look those up, or
sometimes they're given.

34
00:01:07,079 --> 00:01:09,263
It turns out the specific heat of water

35
00:01:09,263 --> 00:01:13,367
is 4,186 joules

36
00:01:13,367 --> 00:01:16,534
per kilogram degree Celsius.

37
00:01:16,534 --> 00:01:17,807
And these units give you an idea

38
00:01:17,807 --> 00:01:19,841
of what the meaning of
the specific heat is.

39
00:01:19,841 --> 00:01:24,334
It's telling you that
water takes 4,186 joules

40
00:01:24,334 --> 00:01:28,079
to heat up one kilogram
by one degree Celsius.

41
00:01:28,079 --> 00:01:29,076
So you can already tell,

42
00:01:29,076 --> 00:01:30,277
this is gonna take a lot of heat.

43
00:01:30,277 --> 00:01:32,370
Water has a very high specific heat.

44
00:01:32,370 --> 00:01:33,974
It can store a lot of heat energy

45
00:01:33,974 --> 00:01:36,236
without raising its temperature by much.

46
00:01:36,236 --> 00:01:37,736
And so let's say the question was

47
00:01:37,736 --> 00:01:40,200
we wanted to get this temperature up to

48
00:01:40,200 --> 00:01:41,867
50 degrees Celsius.

49
00:01:41,867 --> 00:01:44,207
How much heat energy are we going to add

50
00:01:44,207 --> 00:01:46,675
in order to get it to 50 degrees Celsius?

51
00:01:46,675 --> 00:01:48,045
Well, we can come over to here.

52
00:01:48,045 --> 00:01:50,479
The amount of heat that
we're going to need will be

53
00:01:50,479 --> 00:01:52,176
the mass is two kilograms,

54
00:01:52,176 --> 00:01:54,880
so two kilograms times the specific heat

55
00:01:54,880 --> 00:01:57,702
which is 4,186

56
00:01:57,702 --> 00:01:59,714
times the change in temperature.

57
00:01:59,714 --> 00:02:03,380
This really means T final minus T initial.

58
00:02:03,380 --> 00:02:05,039
So what's T final gonna be?

59
00:02:05,039 --> 00:02:07,407
T final is 50 degrees Celsius,

60
00:02:07,407 --> 00:02:08,578
that's where we wanted to get,

61
00:02:08,578 --> 00:02:11,679
minus the initial, it
started at 20 Celsius.

62
00:02:11,679 --> 00:02:13,541
And now we can solve this
for the amount of heat.

63
00:02:13,541 --> 00:02:15,066
And if you multiply all this out

64
00:02:15,066 --> 00:02:20,066
you get 251,160 joules.

65
00:02:20,712 --> 00:02:22,381
That's a lot of heat energy

66
00:02:22,381 --> 00:02:25,771
just to get water to increase
by 30 degrees Celsius.

67
00:02:25,771 --> 00:02:28,075
That's why we often use
water as a heat sink.

68
00:02:28,075 --> 00:02:29,947
You can put a lot of heat in water

69
00:02:29,947 --> 00:02:32,539
and not change the
temperature by all that much.

70
00:02:32,539 --> 00:02:34,238
But that example was pretty simple.

71
00:02:34,238 --> 00:02:35,767
Let's look at a harder one.

72
00:02:35,767 --> 00:02:37,736
Let's say instead of heating the cube

73
00:02:37,736 --> 00:02:39,107
with a fire underneath

74
00:02:39,107 --> 00:02:40,933
we just take a hot piece of metal.

75
00:02:40,933 --> 00:02:44,207
Say we've got a 0.5
kilogram piece of copper

76
00:02:44,207 --> 00:02:45,700
and we drop it into the water.

77
00:02:45,700 --> 00:02:47,102
We've heated up this copper

78
00:02:47,102 --> 00:02:49,679
and now we drop it in the
same container of water.

79
00:02:49,679 --> 00:02:51,713
And we wanna know what equilibrium

80
00:02:51,713 --> 00:02:53,312
temperature will they reach?

81
00:02:53,312 --> 00:02:55,933
The copper is gonna cool down,
the water is gonna heat up.

82
00:02:55,933 --> 00:02:57,671
Eventually they're gonna reach equilibrium

83
00:02:57,671 --> 00:02:58,904
at some temperature.

84
00:02:58,904 --> 00:03:00,377
What temperature will that be?

85
00:03:00,377 --> 00:03:02,000
Well we're gonna need
to know a few things.

86
00:03:02,000 --> 00:03:03,845
I already told you the mass of the copper.

87
00:03:03,845 --> 00:03:06,044
We're gonna need to know
its initial temperature,

88
00:03:06,044 --> 00:03:07,709
so the initial temperature of the copper.

89
00:03:07,709 --> 00:03:09,066
I said we made it really hot.

90
00:03:09,066 --> 00:03:11,407
Let's say it's 90 degrees Celsius.

91
00:03:11,407 --> 00:03:13,673
We'll need to know the
specific heat of copper

92
00:03:13,673 --> 00:03:17,708
and it turns out the specific
heat of copper is 387.

93
00:03:17,708 --> 00:03:18,968
And let's say that water has the same

94
00:03:18,968 --> 00:03:20,566
properties it had initially.

95
00:03:20,566 --> 00:03:22,112
There were two kilograms of it.

96
00:03:22,112 --> 00:03:26,838
The specific heat is
always 4,186 for water.

97
00:03:26,838 --> 00:03:28,678
And let's say it started at a temperature

98
00:03:28,678 --> 00:03:30,800
of 20 degrees Celcius.

99
00:03:30,800 --> 00:03:32,766
So we know the equilibrium temperature,

100
00:03:32,766 --> 00:03:34,639
the temperature at which
these are going to meet

101
00:03:34,639 --> 00:03:36,712
is somewhere between 20 and 90.

102
00:03:36,712 --> 00:03:38,946
We have to figure out
exactly where it's gonna be.

103
00:03:38,946 --> 00:03:41,314
And the trick we're gonna
use is if you think about it

104
00:03:41,314 --> 00:03:42,772
the copper is gonna lose heat,

105
00:03:42,772 --> 00:03:44,107
the water is gonna gain heat,

106
00:03:44,107 --> 00:03:45,466
how do those heats compare?

107
00:03:45,466 --> 00:03:46,677
They've gotta be the same

108
00:03:46,677 --> 00:03:49,644
assuming no heat is being
lost to the environment.

109
00:03:49,644 --> 00:03:51,236
We're gonna assume no heat's lost

110
00:03:51,236 --> 00:03:52,742
so you want this to happen in

111
00:03:52,742 --> 00:03:54,405
what's often called a calorimeter,

112
00:03:54,405 --> 00:03:55,745
something insulated,

113
00:03:55,745 --> 00:03:58,038
something that prevents
any heat from getting out.

114
00:03:58,038 --> 00:03:59,444
And if no heat gets out

115
00:03:59,444 --> 00:04:01,079
then whatever heat the water gains

116
00:04:01,079 --> 00:04:03,939
has to be the same as the
heat lost by the copper.

117
00:04:03,939 --> 00:04:06,945
Basically, if you add up
the heat from the copper

118
00:04:06,945 --> 00:04:10,143
plus the heat from the
water you're gonna get zero

119
00:04:10,143 --> 00:04:11,813
because one of these are gonna be negative

120
00:04:11,813 --> 00:04:13,404
and one of these gonna be positive,

121
00:04:13,404 --> 00:04:15,405
and they're gonna be
the same absolute value.

122
00:04:15,405 --> 00:04:16,476
How do you find these?

123
00:04:16,476 --> 00:04:17,243
We had a formula.

124
00:04:17,243 --> 00:04:21,105
Remember Q equals MC delta
T which I like to remember

125
00:04:21,105 --> 00:04:23,307
because it looks like MCAT,

126
00:04:23,307 --> 00:04:25,866
so MC, this delta looks like an A to me,

127
00:04:25,866 --> 00:04:27,876
so this looks like Q equals MCAT.

128
00:04:27,876 --> 00:04:29,498
So I've got to use the mass of copper

129
00:04:29,498 --> 00:04:32,307
there was 0.5 kilograms of copper

130
00:04:32,307 --> 00:04:35,270
times the specific heat which is 387,

131
00:04:35,270 --> 00:04:37,174
times the change in temperature.

132
00:04:37,174 --> 00:04:38,640
I don't know the final temperature,

133
00:04:38,640 --> 00:04:39,467
that's okay.

134
00:04:39,467 --> 00:04:42,675
I'm going to name my ignorance
and give it a variable here.

135
00:04:42,675 --> 00:04:44,308
I'm gonna call it T final minus,

136
00:04:44,308 --> 00:04:45,611
I don know the initial.

137
00:04:45,611 --> 00:04:47,476
It started at 90 degrees Celsius,

138
00:04:47,476 --> 00:04:49,611
so minus 90 degrees Celsius,

139
00:04:49,611 --> 00:04:52,341
plus the heat gained by the water

140
00:04:52,341 --> 00:04:54,998
which we can use the
same formula for, MCAT.

141
00:04:54,998 --> 00:04:57,100
So the mass is two kilograms.

142
00:04:57,100 --> 00:05:00,344
The specific heat is 4,186.

143
00:05:00,344 --> 00:05:03,370
T final, I still don't know T final,

144
00:05:03,370 --> 00:05:04,914
but I do know the T initial,

145
00:05:04,914 --> 00:05:07,645
the initial temperature
is 20 degrees Celsius.

146
00:05:07,645 --> 00:05:10,800
And I've got to set
this all equal to zero.

147
00:05:10,800 --> 00:05:11,778
Ran out of room there.

148
00:05:11,778 --> 00:05:12,874
Sorry about that.

149
00:05:12,874 --> 00:05:14,340
This looks a little intimidating now

150
00:05:14,340 --> 00:05:15,578
you got this big equation.

151
00:05:15,578 --> 00:05:17,013
You've got your unknown hidden in here.

152
00:05:17,013 --> 00:05:18,246
Is this solvable?

153
00:05:18,246 --> 00:05:19,276
Yeah, it's solvable.

154
00:05:19,276 --> 00:05:20,712
Look, you've only got one unknown.

155
00:05:20,712 --> 00:05:22,273
The unknown is T final.

156
00:05:22,273 --> 00:05:24,533
These are both the same variable.

157
00:05:24,533 --> 00:05:26,202
This is the temperature at which

158
00:05:26,202 --> 00:05:28,012
the water and copper are gonna meet.

159
00:05:28,012 --> 00:05:29,977
This whole term here, this orange term,

160
00:05:29,977 --> 00:05:31,677
is gonna come out to
be some negative number

161
00:05:31,677 --> 00:05:33,710
because the copper is
gonna lose heat energy,

162
00:05:33,710 --> 00:05:36,001
and the water term is
gonna come out positive

163
00:05:36,001 --> 00:05:37,813
because it's gonna gain heat energy.

164
00:05:37,813 --> 00:05:38,975
The two will cancel out.

165
00:05:38,975 --> 00:05:40,140
They'll give you zero.

166
00:05:40,140 --> 00:05:41,911
That's the condition that we're requiring.

167
00:05:41,911 --> 00:05:43,705
We just have to solve for T final

168
00:05:43,705 --> 00:05:45,437
which means we multiply all this out,

169
00:05:45,437 --> 00:05:47,201
combine the T final terms,

170
00:05:47,201 --> 00:05:48,807
and then solve for T final.

171
00:05:48,807 --> 00:05:51,335
First, I'm just gonna
multiply everything out.

172
00:05:51,335 --> 00:05:54,078
I'm gonna combine the T final terms,

173
00:05:54,078 --> 00:05:56,712
and the two terms that don't have T final.

174
00:05:56,712 --> 00:05:59,106
Then I'm gonna move everything
over to the other side.

175
00:05:59,106 --> 00:06:03,943
I solve for T final and I
get 21.58 degree Celsius.

176
00:06:03,943 --> 00:06:05,479
And when you look at this you might think,

177
00:06:05,479 --> 00:06:06,770
"Hey, we must have screwed up."

178
00:06:06,770 --> 00:06:08,707
21.58?

179
00:06:08,707 --> 00:06:10,234
The water started at 20.

180
00:06:10,234 --> 00:06:12,638
It barely increased
its temperature at all.

181
00:06:12,638 --> 00:06:14,378
Yeah, that's what we were saying is that

182
00:06:14,378 --> 00:06:16,745
this specific heat for water is so high

183
00:06:16,745 --> 00:06:18,098
you can add a lot of heat

184
00:06:18,098 --> 00:06:19,706
and it doesn't change
its temperature much.

185
00:06:19,706 --> 00:06:22,675
Note that we could've
added in this container.

186
00:06:22,675 --> 00:06:25,103
This container might
absorb some of that heat,

187
00:06:25,103 --> 00:06:26,808
so we could've had another term over here,

188
00:06:26,808 --> 00:06:28,572
Q of the container,

189
00:06:28,572 --> 00:06:30,178
and taken that into account,

190
00:06:30,178 --> 00:06:32,907
or we could've dropped another
cube of something in here

191
00:06:32,907 --> 00:06:34,514
and we could've added that over here.

192
00:06:34,514 --> 00:06:37,573
If you add up all the Qs
from everything involved

193
00:06:37,573 --> 00:06:39,569
that might gain heat or lose heat

194
00:06:39,569 --> 00:06:41,047
you can set that equal to zero

195
00:06:41,047 --> 00:06:42,646
because if no heat's getting out

196
00:06:42,646 --> 00:06:45,376
that heat just has to be
interchanged within here.

197
00:06:45,376 --> 00:06:47,541
No heat is gonna be created or destroyed.

198
00:06:47,541 --> 00:06:49,400
It just gets transferred amongst the

199
00:06:49,400 --> 00:06:51,378
materials that are interacting.

200
00:06:51,378 --> 00:06:53,937
So this is the key problem solving idea

201
00:06:53,937 --> 00:06:55,937
when you're doing these
specific heat problems.

202
00:06:55,937 --> 00:06:57,244
You set it up with this

203
00:06:57,244 --> 00:06:58,539
and then you solve for the unknown.

204
00:06:58,539 --> 00:07:00,146
In this case it was T final.

205
00:07:00,146 --> 00:07:01,412
Sometimes the thing you won't know

206
00:07:01,412 --> 00:07:02,813
would be the mass of one of them

207
00:07:02,813 --> 00:07:04,279
or the specific heat of one of them

208
00:07:04,279 --> 00:07:06,543
regardless, you solve for
the thing you wanna find.

209
00:07:06,543 --> 00:07:07,867
Let me ask you another question.

210
00:07:07,867 --> 00:07:10,200
Let's say we took the
same amount of water,

211
00:07:10,200 --> 00:07:14,277
two kilograms at a temperature
of 20 degrees Celsius,

212
00:07:14,277 --> 00:07:16,414
but this time I wanna know how much heat

213
00:07:16,414 --> 00:07:18,676
do I have to add in order to boil

214
00:07:18,676 --> 00:07:20,735
all of this water into steam?

215
00:07:20,735 --> 00:07:21,874
Well, the first thing we need to do

216
00:07:21,874 --> 00:07:23,669
is get this to the boiling temperature.

217
00:07:23,669 --> 00:07:25,166
And the boiling temperature of water

218
00:07:25,166 --> 00:07:26,805
is 100 degrees Celsius.

219
00:07:26,805 --> 00:07:29,347
So Q, I first have to use MCAT.

220
00:07:29,347 --> 00:07:30,534
MC delta T.

221
00:07:30,534 --> 00:07:31,813
The mass is two.

222
00:07:31,813 --> 00:07:34,968
The specific heat of water is 4,186.

223
00:07:34,968 --> 00:07:36,310
The change in temperature,

224
00:07:36,310 --> 00:07:38,273
well, the boiling point of water is 100,

225
00:07:38,273 --> 00:07:40,768
so I have to get this
water from 20 to 100.

226
00:07:40,768 --> 00:07:42,580
That means T final is 100.

227
00:07:42,580 --> 00:07:44,309
T initial was 20.

228
00:07:44,309 --> 00:07:46,134
And I get that the heat
that needs to be added

229
00:07:46,134 --> 00:07:48,880
in order to get this to
the boiling temperature is

230
00:07:48,880 --> 00:07:52,973
669,760.

231
00:07:52,973 --> 00:07:54,810
I'm being sloppy with significant figures

232
00:07:54,810 --> 00:07:56,867
but this is the number you
get from that calculation.

233
00:07:56,867 --> 00:07:58,600
But that's not enough to boil it.

234
00:07:58,600 --> 00:08:00,145
That's just the heat required

235
00:08:00,145 --> 00:08:04,012
to get the water up to
100 degrees Celsius.

236
00:08:04,012 --> 00:08:04,904
That's not enough.

237
00:08:04,904 --> 00:08:06,634
If you get water to a 100 degrees Celsius

238
00:08:06,634 --> 00:08:08,637
and let is sit there,
it'll just sit there.

239
00:08:08,637 --> 00:08:09,973
It won't actually boil.

240
00:08:09,973 --> 00:08:11,766
You gotta keep adding heat.

241
00:08:11,766 --> 00:08:14,042
How much more heat are
we gonna have to add?

242
00:08:14,042 --> 00:08:15,900
Once this water gets to 100,

243
00:08:15,900 --> 00:08:18,645
in order to boil all of
this water into steam,

244
00:08:18,645 --> 00:08:20,240
for that you need to know about the heat

245
00:08:20,240 --> 00:08:21,880
of fusion and vaporization.

246
00:08:21,880 --> 00:08:23,547
In this case since it's boiling

247
00:08:23,547 --> 00:08:25,340
the heat of vaporization because

248
00:08:25,340 --> 00:08:26,945
we're turning liquid into vapor.

249
00:08:26,945 --> 00:08:28,909
If we were turning liquid into a solid

250
00:08:28,909 --> 00:08:30,272
it'd be the heat of fusion.

251
00:08:30,272 --> 00:08:32,875
The formula for the heat
of fusion and vaporization

252
00:08:32,875 --> 00:08:34,107
looks like this.

253
00:08:34,107 --> 00:08:35,977
Q, the amount of heat you need to add

254
00:08:35,977 --> 00:08:37,912
in order to change the phase.

255
00:08:37,912 --> 00:08:40,066
This is what happens
when you change phase,

256
00:08:40,066 --> 00:08:41,844
heat of fusion and vaporization.

257
00:08:41,844 --> 00:08:43,107
Specific heat is what happens

258
00:08:43,107 --> 00:08:44,579
when you change temperature.

259
00:08:44,579 --> 00:08:45,911
So this calculation showed us

260
00:08:45,911 --> 00:08:47,976
how much heat we needed to change

261
00:08:47,976 --> 00:08:50,590
the temperature by 80 degrees Celsius.

262
00:08:50,590 --> 00:08:52,145
This calculation is gonna tell us

263
00:08:52,145 --> 00:08:53,471
once we're at a 100

264
00:08:53,471 --> 00:08:54,979
how much heat do we need to add

265
00:08:54,979 --> 00:08:58,307
to change the phase of all
of this water into vapor?

266
00:08:58,307 --> 00:08:59,634
And the formula for the heat of

267
00:08:59,634 --> 00:09:01,334
fusion and vaporization looks like this.

268
00:09:01,334 --> 00:09:02,940
Q equals ML.

269
00:09:02,940 --> 00:09:04,174
M is the mass.

270
00:09:04,174 --> 00:09:06,405
The more mass you have then
the more heat you have to add.

271
00:09:06,405 --> 00:09:09,430
L is the latent heat of
fusion or vaporization.

272
00:09:09,430 --> 00:09:11,476
This is a number similar
to the specific heat,

273
00:09:11,476 --> 00:09:13,033
but instead of telling you how much

274
00:09:13,033 --> 00:09:15,436
heat you have to add in order
to change the temperature,

275
00:09:15,436 --> 00:09:17,245
this is telling you how
much heat you have to add

276
00:09:17,245 --> 00:09:19,214
in order to change the phase.

277
00:09:19,214 --> 00:09:21,036
And it turns out the latent heat

278
00:09:21,036 --> 00:09:24,476
of vaporization for water is huge.

279
00:09:24,476 --> 00:09:29,476
2,260,000 joules per kilogram.

280
00:09:29,976 --> 00:09:33,769
This means it takes 2,260,000 joules

281
00:09:33,769 --> 00:09:35,779
to turn one kilogram of water,

282
00:09:35,779 --> 00:09:37,348
which is at the boiling point,

283
00:09:37,348 --> 00:09:39,398
into one kilogram of vapor.

284
00:09:39,398 --> 00:09:41,537
So, if I'm trying to get
this water from 20 degrees

285
00:09:41,537 --> 00:09:43,002
turned into vapor,

286
00:09:43,002 --> 00:09:45,234
first I need to MC delta T,

287
00:09:45,234 --> 00:09:46,873
get it to 100 degree Celsius.

288
00:09:46,873 --> 00:09:49,540
And then I need to add to
that another amount of heat,

289
00:09:49,540 --> 00:09:51,504
an amount of heat M times L.

290
00:09:51,504 --> 00:09:53,812
The mass is two kilograms of this water.

291
00:09:53,812 --> 00:09:57,870
L for water is 2,260,000.

292
00:09:57,870 --> 00:09:59,476
So I get that it will take

293
00:09:59,476 --> 00:10:03,735
669,760 joules to get the water

294
00:10:03,735 --> 00:10:05,279
up to the boiling point,

295
00:10:05,279 --> 00:10:06,777
and it will take another

296
00:10:06,777 --> 00:10:10,547
4,520,000 joules

297
00:10:10,547 --> 00:10:13,044
to turn that water into vapor

298
00:10:13,044 --> 00:10:14,808
which in total gives

299
00:10:14,808 --> 00:10:19,808
5,189,760 joules

300
00:10:20,078 --> 00:10:21,978
in order to get this from 20

301
00:10:21,978 --> 00:10:24,646
all the way up to two kilograms of vapor.

302
00:10:24,646 --> 00:10:26,378
Now, there's one more thing
that I wanna show you.

303
00:10:26,378 --> 00:10:27,500
Let's get rid of this.

304
00:10:27,500 --> 00:10:28,910
Let's say instead of starting with water

305
00:10:28,910 --> 00:10:30,605
you started with some ice.

306
00:10:30,605 --> 00:10:32,602
Let's say you started
with a big piece of ice,

307
00:10:32,602 --> 00:10:34,773
three kilograms worth of ice.

308
00:10:34,773 --> 00:10:35,978
And it was very cold.

309
00:10:35,978 --> 00:10:37,173
This wasn't just at zero.

310
00:10:37,173 --> 00:10:39,545
Let's say this started
at an initial temperature

311
00:10:39,545 --> 00:10:42,180
of negative 40 degrees Celsius.

312
00:10:42,180 --> 00:10:43,600
The question I wanna know

313
00:10:43,600 --> 00:10:45,941
how much heat would we have to add

314
00:10:45,941 --> 00:10:49,312
in order to turn this
three kilogram block of ice

315
00:10:49,312 --> 00:10:52,709
into three kilograms of water vapor?

316
00:10:52,709 --> 00:10:55,080
But not just get it to
the point of being vapor.

317
00:10:55,080 --> 00:10:57,779
I wanna get it to hotter
than a 100 degrees Celsius.

318
00:10:57,779 --> 00:10:59,701
I wanna get this to a T final

319
00:10:59,701 --> 00:11:02,443
of 160 degrees Celsius.

320
00:11:02,443 --> 00:11:04,175
How much heat would I have to add

321
00:11:04,175 --> 00:11:05,346
in order to do that?

322
00:11:05,346 --> 00:11:07,997
And to track this, to
visualize this a little better,

323
00:11:07,997 --> 00:11:10,181
I'm gonna track what the temperature is

324
00:11:10,181 --> 00:11:13,270
on this vertical axis as a function of

325
00:11:13,270 --> 00:11:16,034
how much heat we've
added into this system.

326
00:11:16,034 --> 00:11:18,871
Let me quickly show
you how not to do this.

327
00:11:18,871 --> 00:11:23,046
A naive approach would say,
"Okay, Q equals MC delta T."

328
00:11:23,046 --> 00:11:26,078
My M is three kilograms.

329
00:11:26,078 --> 00:11:28,109
My C is, well, whatever my C is.

330
00:11:28,109 --> 00:11:29,279
We'll talk about that in a minute.

331
00:11:29,279 --> 00:11:30,168
Delta T.

332
00:11:30,168 --> 00:11:31,841
All right, delta T, my final temperature

333
00:11:31,841 --> 00:11:33,866
is 160 degree Celsius.

334
00:11:33,866 --> 00:11:36,002
My initial temperature is negative 40.

335
00:11:36,002 --> 00:11:37,703
You can't forget the negative sign.

336
00:11:37,703 --> 00:11:39,900
And I just plug in my specific heat

337
00:11:39,900 --> 00:11:41,776
and I find my value, this is wrong.

338
00:11:41,776 --> 00:11:43,536
You cannot do it this way.

339
00:11:43,536 --> 00:11:45,611
For one what specific
heat are you gonna put in?

340
00:11:45,611 --> 00:11:47,202
The specific heat of water?

341
00:11:47,202 --> 00:11:48,434
The specific heat of ice?

342
00:11:48,434 --> 00:11:49,698
The specific heat of vapor?

343
00:11:49,698 --> 00:11:51,077
They all have different specific heats.

344
00:11:51,077 --> 00:11:53,613
And for two there's
gonna be phase changes.

345
00:11:53,613 --> 00:11:55,267
First the ice turns into water

346
00:11:55,267 --> 00:11:57,513
and then at some later point
the water turns into vapor.

347
00:11:57,513 --> 00:12:00,977
You can't just neglect and
gloss over those phase changes.

348
00:12:00,977 --> 00:12:02,614
So this is not how you do it.

349
00:12:02,614 --> 00:12:03,847
Here's what you should do.

350
00:12:03,847 --> 00:12:06,608
We're gonna start at
negative 40 degrees Celsius.

351
00:12:06,608 --> 00:12:08,668
I know this is above the axis over here

352
00:12:08,668 --> 00:12:10,133
but just pretend like this isn't

353
00:12:10,133 --> 00:12:12,300
the zero point for this vertical axis.

354
00:12:12,300 --> 00:12:14,237
And we're gonna add heat
and that's gonna bring

355
00:12:14,237 --> 00:12:17,213
this temperature up to
zero degrees Celsius,

356
00:12:17,213 --> 00:12:18,679
and we have to stop there.

357
00:12:18,679 --> 00:12:20,538
We have to pause at zero degrees Celsius

358
00:12:20,538 --> 00:12:22,177
because we have to pause every time

359
00:12:22,177 --> 00:12:23,672
there's gonna be a phase change.

360
00:12:23,672 --> 00:12:27,369
So how much heat does this
correspond to right here?

361
00:12:27,369 --> 00:12:29,010
We can use MC delta T.

362
00:12:29,010 --> 00:12:30,600
M is three kilograms.

363
00:12:30,600 --> 00:12:34,676
The specific heat of ice is about 2,090

364
00:12:34,676 --> 00:12:36,510
and the final temperature is zero,

365
00:12:36,510 --> 00:12:37,806
our initial temperature,

366
00:12:37,806 --> 00:12:40,879
so T final minus our
initial is negative 40.

367
00:12:40,879 --> 00:12:42,411
Don't forget that negative sign.

368
00:12:42,411 --> 00:12:45,045
And if you calculate that
you get that amount of heat.

369
00:12:45,045 --> 00:12:48,867
It's 250,800 joules.

370
00:12:48,867 --> 00:12:51,207
But that's just getting the
ice up to the melting point.

371
00:12:51,207 --> 00:12:52,535
Now we need to melt it.

372
00:12:52,535 --> 00:12:54,639
What's this graph gonna
look like while it melts?

373
00:12:54,639 --> 00:12:55,933
Well the temperature of the ice

374
00:12:55,933 --> 00:12:58,406
is gonna stay constant
while this thing melts.

375
00:12:58,406 --> 00:13:00,078
As you're melting the ice cube

376
00:13:00,078 --> 00:13:01,573
the temperature doesn't change.

377
00:13:01,573 --> 00:13:04,038
All that energy is going
into breaking those bonds

378
00:13:04,038 --> 00:13:05,941
and turning this ice into water.

379
00:13:05,941 --> 00:13:08,300
So, how much heat does this correspond to?

380
00:13:08,300 --> 00:13:11,635
It's a phase change so
you gotta use Q equals ML.

381
00:13:11,635 --> 00:13:13,772
The M is three kilograms.

382
00:13:13,772 --> 00:13:16,144
The latent heat, we
can't use the latent heat

383
00:13:16,144 --> 00:13:17,380
of vaporization.

384
00:13:17,380 --> 00:13:19,346
This is a solid turning into a liquid.

385
00:13:19,346 --> 00:13:21,371
That's latent heat of fusion that we need,

386
00:13:21,371 --> 00:13:23,133
and the latent heat of fusion for water

387
00:13:23,133 --> 00:13:27,268
is about 333,000 joules per kilogram

388
00:13:27,268 --> 00:13:31,743
which gives you 999,000 joules of heat

389
00:13:31,743 --> 00:13:34,510
in order to turn this ice
at zero degree Celsius

390
00:13:34,510 --> 00:13:36,907
into water at zero degrees Celsius.

391
00:13:36,907 --> 00:13:38,205
Now you see how this works.

392
00:13:38,205 --> 00:13:40,776
We've got to take this
water at zero up to what?

393
00:13:40,776 --> 00:13:42,165
Not up to 160.

394
00:13:42,165 --> 00:13:45,178
Up to 100 degrees Celsius because that's

395
00:13:45,178 --> 00:13:46,678
where it's gonna turn into steam.

396
00:13:46,678 --> 00:13:48,513
And whenever there's a
phase change you gotta pause

397
00:13:48,513 --> 00:13:50,938
because your specific heat
is gonna change its value.

398
00:13:50,938 --> 00:13:53,475
So this Q that is required right here

399
00:13:53,475 --> 00:13:55,209
we can do MC delta T,

400
00:13:55,209 --> 00:13:58,432
specific heat of water is 4,186.

401
00:13:58,432 --> 00:14:02,540
Delta T is, final is 100, initial is zero.

402
00:14:02,540 --> 00:14:07,374
We get 1,255,800 joules.

403
00:14:07,374 --> 00:14:09,637
Now we've got to turn
that water into steam.

404
00:14:09,637 --> 00:14:11,238
How much heat is that gonna take?

405
00:14:11,238 --> 00:14:12,447
That's a phase change.

406
00:14:12,447 --> 00:14:15,575
So we've gotta use ML, the
mass is three kilograms.

407
00:14:15,575 --> 00:14:17,903
The latent heat of vaporization is

408
00:14:17,903 --> 00:14:21,566
2,260,000 which means we need

409
00:14:21,566 --> 00:14:25,813
6,780,000 joules in order to turn

410
00:14:25,813 --> 00:14:27,580
this water into vapor,

411
00:14:27,580 --> 00:14:29,247
and we've got one more step to go.

412
00:14:29,247 --> 00:14:31,180
We've got to turn this vapor at a 100

413
00:14:31,180 --> 00:14:33,181
into vapor at a 160.

414
00:14:33,181 --> 00:14:35,641
We've got to do one more MC delta T,

415
00:14:35,641 --> 00:14:38,012
the mass of the steam is three kilograms.

416
00:14:38,012 --> 00:14:41,147
The specific heat of steam is about 2,010.

417
00:14:41,147 --> 00:14:43,333
The final temperature is 160.

418
00:14:43,333 --> 00:14:45,736
The initial temperature was 100

419
00:14:45,736 --> 00:14:50,311
which gives us 361,800 joules.

420
00:14:50,311 --> 00:14:52,042
That's how much heat it would take

421
00:14:52,042 --> 00:14:53,879
to get this ice cube at negative 40

422
00:14:53,879 --> 00:00:00,000
into vapor at 160.