Question

After eating a hearty stew you cooked over a campfire with your 0.250-kg aluminum-alloy pot, you place the pot in a plastic bucket containing 1.00 kg of water. The water's temperature increases 1.00°C and the temperature of the pot decreases 17.5°C. Disregarding energy transfer as heat to the surrounding air and bucket, what is the specific heat capacity of the pot?

Answer 1

The specific heat capacity of the pot is -956.71 J/kg°C.

Step-by-step explanation:

In order to find the specific heat capacity of the pot, we first need to calculate the heat transfer for the water and the pot separately. The heat transfer for the water can be calculated using the equation Qw = mwcwΔT, where Qw is the heat transfer, mw is the mass of water, cw is the specific heat capacity of water, and ΔT is the change in temperature. Plugging in the values, we get Qw = (1.00 kg)(4186 J/kg°C)(1.00°C) = 4186 J.

To calculate the heat transfer for the pot, we use the equation Qp = mpcpΔT, where Qp is the heat transfer, mp is the mass of the pot, cp is the specific heat capacity of the pot, and ΔT is the change in temperature. Plugging in the values, we get Qp = (0.250 kg)(cp)(-17.5°C).

Since the heat transfer for the pot is equal and opposite to the heat transfer for the water, we can set Qw = -Qp. Thus, 4186 J = (0.250 kg)(cp)(-17.5°C).

Solving for cp, we get cp = -4186 J / (0.250 kg)(-17.5°C) = 956.71 J/kg°C.