A metal ball of specific gravity 4.5 and specific heat 0.1 cal/gm-°C is placed on a large slab of ice at 0°C. When ball’s temperature become 0°C then half of the ball sinks in the ice. The initial temperature of the ball is:- (Latent heat capacity of ice = 80 cal/g, specific gravity of ice = 0.9)
Text Solution
Verified by ExpertsThe correct answer is:
C
V × 4.5 × 10 3 × 0.1 × T = 
× 0.9 × 10 3 × 80
T = 
= 80ºC.
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