What is the amount of heat energy required to convert ice of mass 20 Kg at $- 4^{0} C$ to water at $20^{0} C$ ? Use the relevant option for calculation.
a) Latent heat of fusion of ice 3.34 x $10^{5}$ J / Kg.
b) Specific heat capacity of water is 4180 J / Kg / K
c) Specific heat of ice 2093 J / Kg / K

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Solution

Given: ice of mass 20 Kg at $- 4^{0} C$ to water at $20^{0} C$
To find: the amount of heat energy required to convert ice to water
Solution:
As per the given condition,
Latent heat of fusion of ice, $L = 3.34 \times 10^{5} J / K g .$
Specific heat capacity of water, $s = 4180 J / K g / K$
Specific heat of ice, $c = 2093 J / K g / K$
Mass of ice, $m = 20 k g$
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 $0^{\circ} C$ to $20^{\circ} C$
$q_{t o t a l} = q_{1} + q_{2} + q_{3} ⟹ q_{t o t a l} = m c (Δ T) + m L + m s (Δ T) ⟹ q_{t o t a l} = 20 \times 2093 \times (0 - (- 4)) + 20 \times 3.34 \times 10^{5} + 20 \times 4180 (20 - 0) ⟹ q_{t o t a l} = 167440 + 66.8 \times 10^{5} + 1672000 ⟹ q_{t o t a l} = 8519440 J = 8.52 \times 10^{3} k J$
is the quantity of heat is required to transform the ice into water

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