A metal ball of mass 0.1 kg is heated upto 500°C and dropped into a vessel of heat capacity 800 JK –1 and containing 0.5 kg water. The initial temperature of water and vessel is 30°C. What is the approximate percentage increment in the temperature of the water? [Specific Heat Capacities of water and metal are, respectively, 4200 Jkg –1 K –1 and 400Jkg –1 K –1 ]
Text Solution
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Considering the subscript for ball as 'b', for water as 'w' and for container as 'c' and applying
principle of calorimetry (assuming final temperature = T°C)
m b s b (500 – T) = m w s w (T – 30) + m c s c (T – 30)
∴ 0.1 × 400 (500 – T) = 0.5 × 4200 (T – 30) + 800 (T – 30)
∴ 20000 – 40T = 2100T – 63000 + 800T – 24000
∴ 2940T = 107000
∴ T = 
= 36.4 °C
% rise in temperature = 
×100% 
20%
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