Question

A piece of copper of mass 300g at a temperature of 130 degree Celsius is quickly transferred to a vessel of negligible thermal capacity containing 250g of water at 30 degrees Celsius. Calculate the final temperature of the system.

Answer 1

The copper will cooled, while the water will heated. We can write the heat balance equation:

$$m_cc_c(t_c-T)=m_wc_w(T-t_w),$$ $$T=\dfrac{m_cc_ct_c+m_wc_wt_w}{m_wc_w+m_cc_c},$$

T=\dfrac{0.3\ kg\cdot 390\ \dfrac{J}{kg\cdot \!^{\circ}C}\cdot130\ ^{\circ}C+0.25\ kg\cdot4200\ \dfrac{J}{kg\cdot \!^{\circ}C}\cdot30\ ^{\circ}C}{0.25\ kg\cdot4200\ \dfrac{J}{kg\cdot \!^{\circ}C}+0.3\ kg\cdot 390\ \dfrac{J}{kg\cdot \!^{\circ}C}}=40\ ^{\circ}C.