Question

A 100g sample of Aluminum ( C= 0.90J/gC)at 120°C and a 150g sample of Iron metal (C=0.45J/C)at 90°C were placed in a container of 80g of water at 15°C. What is the final temperature of the mixture?

Can you show me the solution?

Answer 1

The equation of the thermal equilibrium in this case is:

$$c_am_a(T_a-T_0) + c_im_i(T_i-T_0) = c_wm_w(T_0-T_w)$$

where $c_a = 0.9J/g/C, c_i = 0.45J/g/C, c_w=4.2J/g/C$ are the specific heat capacities of the corresponding materials, $m_a = 100g, m_i = 150g, m_w=80g$ are the masses, and $T_a = 120\degree C, T_i = 90\degree C, T_w=15\degree C$ are the initial temperatures. $T_0$ is the final temperature. Expressing $T_0$, obtain:

$$T_0 = \dfrac{c_am_aT_a + c_im_iT_i+c_wm_wT_w}{c_am_a + c_im_i+ c_wm_w}$$

Substituting numbers, get:

$$T_0 = \dfrac{0.9\cdot 100\cdot 120 +0.45\cdot 150\cdot 90+80\cdot 4.2\cdot 15}{0.9\cdot 100 +0.45\cdot 150+80\cdot 4.2} \approx 44\degree C$$

Answer. $44\degree C$.