Question

A 829$829$ g copper pan is removed from the stove at a temperature of 150 °C. It is then stacked on top of a 611$611$ g aluminum sheet at room temperature, 25 °C. The specific heat capacity for copper is 0.387 Jg·K$\frac{\text{J}}{\text{g·K}}$ and that for aluminum is 0.904 Jg·K$\frac{\text{J}}{\text{g·K}}$. Assume no heat is gained or lost to the surroundings.

What is the final temperature ( 𝑇f ) of the two objects after thermal equilibrium is reached?

Answer 1

SOLUTION :

Aș per energy conservation, heat lost by copper pan is equal to heat gained by 

The aluminium sheet.

Let final temperature of pan and sheet be T º K .

So,

Heat lost by copper pan = Heat gained by the aluminium sheet

=> m coper*s copper*Temperature loss = m aluminium*s aluminium*Temp. gain.

=> 829 g * 0.387 J/(g.K) * ((150+273) - T) = 611 * 0.904 J/(g.K) * (T - (25+273))

=> 829 * 0.387 * (423 - T)  Joules = 611 * 0.904 * (T - 298)

=> 320.823 * (423 - T) = 552.344 * (T - 298)

=> (552.344 + 320.823) T = 320.823 * 423 + 552.344 * 298

=> 873.167 T = 300306.641

=> T = 300306.641 / 873.167

=> T = 343.93 ºK

=> T = 343 - 93 - 273 = 70.93 ºC 

So, final temperature will be 70.93 ºC (ANSWER).