Question

Ans. (a)
NUMERICALS
1. Calculate the amount of heat required to convert 2.5 kg of ice at 0°C into water without change of
temperature. (Latent heat of fusion of ice = 3.34 x 10 J/kg)
​

Answer 1

Answer:

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Explanation:

As per the given condition,

Latent heat of fusion of ice, L=3.34×10

5

J/Kg.

Specific heat capacity of water, s=4180J/Kg/K

Specific heat of ice, c=2093J/Kg/K

Mass of ice, m=20kg

The total heat required will be sum of heat energy needed to change the temperature of ice from –4°C to 0°C plus heat required to undergo a solid to liquid phase change at its melting point plus heat required to heat the sample from 0

∘

C to 20

∘

C

q

total

=q

1

+q

2

+q

3

⟹q

total

=mc(ΔT)+mL+ms(ΔT)

⟹q

total

=20×2093×(0−(−4))+20×3.34×10

5

+20×4180(20−0)

⟹q

total

=167440+66.8×10

5

+1672000

⟹q

total

=8519440J=8.52×10

3

kJ

is the quantity of heat is required to transform the ice into water