JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Heat & Thermodynamics

A piece of ice (heat capacity

= 2100 J kg^{- 1}^{\circ} C^{- 1}

and latent heat

= 3.36 \times 1 0^{5} J kg^{- 1}

) of mass m grams is at -5°C at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the ice-water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heat exchange in the process, the value of m is

Answer: 8

Solution

The heat supplied is used for two purposes:

Raising the temperature of m grams of ice from -5°C to 0°C
Melting 1 gram of ice at 0°C

Heat required to raise temperature:

Q_{1} = m \times c \times Δ T = m \times 2100 \times 5 J = 10500 m J

(where m is in kg)

Heat required to melt 1 gram of ice:

Q_{2} = 1 \times 1 0^{- 3} \times 3.36 \times 1 0^{5} = 336 J

Total heat supplied:

420 = 10500 m \times 1 0^{- 3} + 336

420 = 10.5 m + 336

84 = 10.5 m

m = 8 grams

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Integer answer

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JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Heat & Thermodynamics

Solution

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