Question

A 20 gm piece of metal is heated to 150 c and then dropped into an aluminum calorimeter of mass 500gm containing 500gm of water initially at 25 c find the final equilibrium temperature of the system if the specific heat of the metal is 128.100j/kg-k while the specific heat of water is4200j/kg-k

Answer 1

Heat lost = Heat gained

$m_hs_hΔT_h = m_cs_cΔT_c \\ m_h = 0.02 \;kg \\ m_c = 0.5 \;kg \\ 0.02 \times 128.1 (150 -T_f) = 0.5 \times 4200 (T_f -25) \\ 384.3 -2.562T_f = 2100T_f -52500 \\ 52884.3 = 2102.562T_f \\ T_f = \frac{52884.3}{2102.562} = 25.15 \;C$

Answer: 25.15 °C