First slide

Heat capacity

Question

A metal piece weighing 15 g is heated to $100^{0} C$ and the immersed in a mixture of ice and water in thermal equilibrium. The volume of the mixture is found to be reduced by 0.15 $c m^{3}$ with the temperature of the mixture remaining constant. Find the specific heat of the metal. Given specific gravity of ice = 0.92, specific gravity of water at $0^{0} C$ = 1.0, latent heat of fusion of ice = $80 c a l g^{- 1}$

Easy

A

$0.092 c a l / g^{0} C$
B

$0.124 c a l / g^{0} C$
C

$0.162 c a l / g^{0} C$
D

$0.242 c a l / g^{0} C$

Solution

$\frac{m}{ρ_{i c e}} - \frac{m}{ρ_{H_{2} O}} = 0.15 \Rightarrow m = \frac{0.15 \times 0.92}{0.08}$
$H_{l o s t} = H_{g a i n e d} \Rightarrow 15 \times S \times 100 = \frac{0.15 \times 0.92}{0.08} \times 80 \Rightarrow S = 0.092 c a l / g^{0} C$

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