Question

A cooper cube whose mass mc 100 g is heated to 300 oC. ) Then the cube is dropped intro glass baker whose heat capacity Cb 200 J/K, and containing water whose heat capacity CW 800 J/K at initial temperature 25 oC then the common final temperature is ( the specific heat of cooper 400 J/Kg.k

Answer 1

The heat lost by the copper cube

$Q_{lost} = cm \Delta T=400 \frac{J}{kg\times K}\times 0.1 kg \times (T_{initial}-T_{final})$

$T_{initial} (K) = 273 + 300 = 573 K$

$Q_{lost} = 400 \frac{J}{kg\times K}\times 0.1 kg \times (573 K - T_{final})$

The heat gained by glass baker

$Q_{gl.baker}= C_{b}\times \Delta T = C_{b}\times (T_{final}- T_{initial})$

$T_{initial}(K)= 273+25 = 298 K$

$Q_{gl.baker} = 200 \frac{J}{K} \times (T_{final}-298 K)$

The heat gained by water

$Q_{water} = C_w\times \Delta T = 800 \frac{J}{K}(T_{final}-298K)$

As $Q_{lost} = Q _{gained}$ then

$Q_{lost} = Q_{gl.baker} + Q_{water}$

$400 \frac{J}{kg\times K}\times 0.1 kg\times (573K - T_{final})= 200\frac{J}{K}\times(T_{final}-298 K)+800 \frac {J}{K}\times (T_{final}-298 K)$

Solve for $T_{final}$

$T_{final} = 309 K$

$T_{final}(C ^\circ )= 309-273 = 36 C^\circ$