Section 5.1Temperature


1. (Exercises 1 and 2) On a very cold day in Alaska the thermometer read -47° F. What is this temperature on the Celsius scale? What is this temperature in kelvins?

This is a straight conversion problem, which requires two steps as shown below.

T_C = 5/9 (-47° - 32) = 5/9 (-79°) = -44° C

T_K = T_C + 273 = -44° + 273 = 229 K

2. (Exercises 5 and 6) One of the newest superconductor materials becomes superconducting at a temperature of about 92 K. What is this temperature in ° C and in ° F?

Again two steps of conversion are needed.

T_C = T_K - 273 = 92 - 273 = -181°C

T_F = 9/5 (T_C) + 32 = 9/5 (-181) + 32 = -326 + 32 = -294°F

Section 5.2Heat


3. (Exercises 9 and 10) To bake a cake, about 750 kcal of heat must be supplied by the oven. If the cake takes 15 minutes to bake, how many watts of power must the stove produce?

First, we need to find the number of seconds in 15 minutes.

15 min (60 s / min) = 900 s

You must remember that heat is a form of work and so the number of kilocalories can be directly substituted for the work done to bake the cake.

Then the power can be determined using its defining relationship involving the work done divided by the time.

P = W / t = 750 kcal / 900 s = 0.833 kcal (1000 cal/kcal) = 833 cal/s

Since watt is a unit in the SI, we must convert calories/second to joules/second.

833 cal/s (4.186 J / cal) = 3500 J/s = 3500 W

Section 5.3Specific Heat and Latent Heat


4. (Exercises 15 and 16) When you place an ice tray in your freezer, you must cool the water from room temperature (20°C) to the freezing point, convert the water to ice, and then cool the ice to -10°, where it remains until you use it. If an ice tray holds 375 g of water, how much heat must be removed from the water to cool and freeze it?

This problem must be done in 3 steps. First, we must cool the water from 20°to 0° (the freezing point), then freeze the water, and finally cool the ice to -10°C.

H₁ = m c_w T_w = 0.375 kg (4186 J/kg-°C)(0°C - 20°C) = -31400 J

H₂ = m L_F = 0.375 kg (-335000 J/kg) = -125600 J

H₃ = m c_IT_I = 0.375 kg (2100 J/kg-°)(-10°C - 0°C) = -7880 J

Note that all of these are (-) because heat has been removed in each case. The total heat removed is then:

H_T = H₁ + H₂ + H₃ = (-31400 J) + (-125600 J) + (-7880 J) = -165,000 J

Section 5.6The Kinetic Theory of Gases


5. (Exercises 21 and 22) A cylinder of gas whose volume remains constant is reduced from room temperature (20°C) to the freezing point of water (0°C). What is the pressure of the gas relative to its initial pressure?

Since the volume remains constant, the ideal gas law can be simplified by canceling the volume from each side, leaving only the pressure and temperature variables in the equation.

p₁V₁/T₁ = p₂V₂/T₂ becomes p₁/T₁ = p₂/T₂

Now solving for the unknown quantity p₂:

p₂ = (T₂/T₁)₁ and using the Kelvin temperatures as required

p₂ = [(0° + 273)/(20° + 273)]p₁ = (273/293)p₁ = 0.93 p₁

In this problem the pressure decreases, but only by a small amount, about a 7% decrease from its original value.

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