Find the quantity of heat needed to melt 100g of ice at -10°C into the water at 10°C.
(39900 J)
(Note: Specific heat of ice is $2100 J〖kg〗^{-1} K^{-1}$, the specific heat of water is $4200 J〖kg〗^{-1} K^{-1}$, Latent heat of fusion of ice is 336000 J$〖kg〗^{-1})$.
Difficulty: Medium
Solution: Mass of ice = m = 100 g = $\frac{100}{1000}$ = 0.1 kg
Specific heat of ice = c1 = $2100 J〖kg〗^{-1} K^{-1}$
Latent heat of fusion of ice = L = $336000 J〖kg〗^{-1}$
Specific heat of water = c = $4200 J〖kg〗^{-1} K^{-1}$
Quantity of heat required = Q =?
Case I:
Heat gained by ice from -10°C to 0°C
Q1 = $mc \triangle T$
Q1 = $0.1\times 2100 \times10$ = 2100 J
Case II:
Heat required for ice to melt = Q2 = mL
= $0.1\times 336000$
Q2 = 33600 J
Case III:
The heat is required to raise the temperature of water from 0°C to 10°C
Q3= $mc \triangle T$
Q3 = $0.1\times 4200 \times 10$ = 4200 J
Total heat required = Q = Q1 + Q2 + Q3
Q = 2100 + 33600 + 4200
Q = 39900 J
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