Homework 13
Practice Problems, Chapter 16
44, 45, 46, 48, 49, 51, 52, 53, 54, 55, 57, 58, 59, 74, 76, 81
Practice Problems, Chapter 17
57, 58, 60, 61, 66, 67, 68, 69, 70, 71
Graded Problems Due Tuesday, 11/17/15
Problem 1
A cube of ice of 100 grams at –30 °C is placed into an insulated cup with 200 grams of water at 20
°C.
a. What is the final temperature of the mixture?
b. What is the mass of water in the mixture?
Problem 2
It is said that sun light arriving at the Earth amounts to 1,000 W/m2 of thermal radiation. This is
1,000 J/s of heat flow rate for every square meter of area out here where we are from the sun.
a. What is the total radiative heat flow rate away from the sun? This is in J/s or W.
b. What is the temperature of the surface of the sun if it were a perfect emitter with an emissivity
value of 1?
page 1
Solution 1
When the ice is placed into the water, two possibilities can happen. First, the ice can all reach 0
°C or the water can all reach 0 °C. The ice requires this much heat.
Qice to 0 = mcice ΔT = (0.1 kg)(2, 050 J/kg ⋅ °C )(30 °C ) = 6,150 J
The water can give up this much heat.
Qwater to 0 = mcwater ΔT = (0.2 kg)(4,186 J/kg ⋅ °C )(−20 °C ) = −16,744 J
This means the ice can all goes to 0 °C. When it does, we have 100 grams of ice at 0 °C and
water at this mew temperature when it gives up 6,150 J of heat.
−6,150 J = mcwater ΔT = (0.2 kg)(4,186 J/kg ⋅ °C )(Tf − 20 °C ) ⇒ Tf = 12.654 °C
Next, these two possibilities can happen. First, all of the ice can melt. Second, all of the water
can reach 0 °C. The ice requires this much heat.
Qice melt = Δm ⋅ Lf = (0.1 kg)(33.5 ×104 J/kg) = 33,500 J
The water has this much heat left to give.
Qwater to 0 = −10,594 J
This means the water will all go to water at 0 °C. When this happens, this much ice will melt. All
of the heat from the water will go to melt some the ice.
+10,594 = Δm ⋅ Lf = m(33.5 ×104 J/kg) ⇒ Δm = 0.031624 kg = 32 g
Now, everything is at 0 °C so the process stops. The final temperature of the mixture is 0 °C. The
total amount of water is 232 grams and the ice remaining is 68 grams.
page 2
Solution 2
The power at Earth’s distance form the sun is 1,000 W/m2. The total surface area where this
applies is the surface of the sphere at Earth’s distance form the sun. This distance is 150 billion
meters.
Ptotal = P per unit area Atotal = 1, 000
W
m2
2
(4πRearth
) = 103
orbit
W
m2
⋅ 4π ⋅ (150 ×109 m)2
Ptotal = 2.83 ×1026 W
This is radiated due to the temperature of the sun. The radius of the sun is 0.70 billion meters.
Ptotal = εσAT 4 = (1)(5.67 ×10−8
W
m 2 ⋅K 4
2
)(4πRsun
)T 4 = 3.49 ×1011
W
K4
⋅T 4
The temperature of the surface of the sun is
2.83 ×1026 W = 3.49 ×1011
W
K4
⋅T 4
⇒ T = 5, 340 K
page 3