CalorimetryHard
Question
A metal ball of specific gravity 4.5 and specific heat 0.1 cal/gm-oC is placed on a large slab of ice at 0oC. 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)
(Latent heat capacity of ice = 80 cal/g, specific gravity of ice = 0.9)
Options
A.100oC
B.90oC
C.80oC
D.70oC
Solution
V × 4.5 × 103 × 0.1 × T = 
× 0.9 × 103 × 80
T =
= 80ºC.
× 0.9 × 103 × 80T =
= 80ºC.Create a free account to view solution
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