Question

1. You brew a mug of coffee, but it is too hot to drink! You measure the temperature of the mug of coffee and find it is 180°F. You could cool it with an ice cube, but that would dilute it. Instead, you take a cold chunk of aluminum that you just happen to have in the freezer, place it in the mug of coffee, and wait until the temperature stops changing.

(a) When you drop this 300 gram chunk of aluminum into the mug of coffee (essentially water, with a mass of 400 grams), the final temperature of the coffee is a pleasing, drinkable 150°F. What was the initial temperature of the chunk of aluminum when you took it out of the freezer? You may ignore any energy transfer to or from the mug, or the surroundings.

(b) If you used 300 grams of copper (at the same initial temperature you just calculated) instead of aluminum, then repeated the experiment as outlined above, which would be true? Explain.

A. The final temperature of the coffee would be greater than 150°F.

B. The final temperature of the coffee would be less than 150°F.

C. The final temperature would be equal to 150°F.

(c) Refer back to the initial experiment in part (a). Assume the mug and coffee start at the same initial temperature of 180 °F. If the energy transfer to or from the mug were considered, which of the following would be true? (Still ignore energy transfer to or from the surroundings.) Explain.

A. The calculated final temperature would be greater than 150°F.

B. The calculated final temperature would be less than 150°F.

C. The calculated final temperature would be equal to 150°F.

Answer 1

Using energy conservation: (when energy transfered to or from the mug is ignored)

Heat gained by aluminum = Heat released by coffee

Q1 = Q2

ma*Ca*dT1 = mc*Cc*dT2

Let, Initial temperature of aluminum chunk = T

Initial temperature of coffee = 180 F = 82.22 C

Final temperature of coffee and Aluminum = 150 F = 65.56 C

T(C) = [T(F) - 32]*(5/9)

dT1 = Tf - Ti = 65.56 - T

dT2 = 82.22 - 65.56 = 16.66 C

ma = mass of aluminum cup = 300 g = 0.300 kg

mc = mass of coffee = 400 gm = 0.400 kg

Ca = Specific heat capacity of aluminum cup = 900 J/kg-C

Cw = Specific heat capacity of coffee = 4186 J/kg-C (Same as water)

Now using given values:

0.300*900*(65.56 - T) = 0.400*4186*16.66

T = (0.300*900*65.56 - 0.400*4186*16.66)/(0.300*900)

T = -37.76 C = -35.97 F = Initial temperature of aluminum chunk

Part B.

We know that Specific heat capacity of copper cup = 387 J/kg-C

Now this copper have same initial temperature as calculated above, then

Heat gained by copper = Heat released by coffee

Q1 = Q2

m1*C1*dT1 = mc*Cc*dT2

Suppose final temperature of system now is Tf, then

0.300*387*(Tf - (-37.76)) = 0.400*4186*(82.22 - Tf)

Tf = (0.400*4186*82.22 - 0.300*387*37.76)/(0.300*387 + 0.400*4186)

Tf = 74.44 C = 166 F = Final temperature of system of coffee + copper cup

The final temperature of the coffee would be greater than 150°F

Remember: since we do not need exactly final temperagture of system, so we can deduce this simply using the fact that since copper have very low specific heat capacity than aluminum, So for same amount of copper and aluminum, energy absorbed by copper will be lower, So that means coffee will released lower amount of energy and it's final temperature will be higher than 150 F, which was in case of aluminum chunk.

Part C.

Now when specific heat capacity of mug was not ignored than in that case:

energy gained by aluminum chunk = energy released by coffee + energy released by mug

So as we can see that when energy released by mug is considered than energy released will increase, which means now aluminum chunk have more energy to absorb, So it's final temperature will be higher than 150 F

The calculated final temperature would be greater than 150°F

Let me know if you've any query.