A ball of copper of specific heat capacity $400 \; J \; / Kg\; K$ weighting $400 \; gm$ is transferred from a furnace to a copper calorimeter of mass 300 gm and containing $1 Kg$ of water at $20^ \circ \; C$. The temperature of water rises to $50^ \circ \; C$. What is the initial temperature of the ball?

Given,

Mass of copper ball $(m_b) = 400 \; gm = 0.4 \; kg$


Mass of calorimeter $(m_c) = 300 \; gm = 0.3 \;kg$

Mass of water $(m_w) = 1 \; kg$

Initial temp of water and calorimeter $(θ_2) = 20^{\circ}C$

Final temp $(θ) = 50^{\circ}C $

Specific heat capacity of copper ball $(S_c) = 400 \; J / Kg K $

Specific heat capacity of water $(S_w) = 4200 \; J / Kg C $

Temperature of ball $(θ_1) = ?$

From the principle of calorimetry,

Heat loss = Heat gain

$m_b \; S_c (\theta_1 - \theta) = (m_c \; S_c + m_w \; S_w) \; (\theta - \theta_2)$

or, $0.4 * 400 (\theta_1 - 50) = 0.3 * 400 * + 1 * 4200) \; (50 - 20)$

or, $160 (\theta_1 - 50) = 129600$

or, $\theta_1 = 810 + 50 = 860^{\circ}C$

$\therefore$ Temperature of ball is $860^{\circ}C$