Question

A 10. g cube of copper at a temperature T1 is placed in an insulated cup containing 10. g of water at a temperature T2. If T1 > T2, which of the following is true of the system when it has attained thermal equilibrium? (The specific heat of copper is 0.385 J/(g C).)
a. The temperature of the copper changed more than the temperature of the water.
b. The temperature of the water changed more than the temperature of the copper.
c. The temperature of the water and the copper changed by the same amount.
d. The relative temperature changes of the copper and the water cannot be determined without knowing T1 and T2

Answer 1

Answer:

a. The temperature of the copper changed more than the temperature of the water.

Explanation:

Because we're only considering the isolated system cube-water, the heat of the system should be constant, that implies the heat the cube loses is equal the heat the water gains (because by zero law of thermodynamics heat (Q) flows from hot body to cold body until reach thermal equilibrium and T1>T2). So:

$Q_{cube}=Q_{water}$ (1)

But Q is related with mass (m), specific heat (c) and changes in temperature ($\varDelta T$)in the next way:

$Q=cm\varDelta T$(2)

Using (2) on (1):

$c_{cooper}*m_{cooper}*\varDelta T_{cooper}=c_{water}*m_{waterer}*\varDelta T_{water}$

$(10g)(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(10g)(4.186 \frac{J}{g\,C})(\varDelta T_{water})$

$(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(4.186 \frac{J}{g\,C})(\varDelta T_{water})$

Because we have an equality and 0.385 < 4.186 then $\varDelta T_{cooper}>\varDelta T_{waterer}$ to conserve the equality