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10g of ice at 0℃ absorbs 5460 J of heat energy to melt and change to water at 50℃. Calculate the specific latent heat of fusion of ice. Specific heat capacity of water is 4200 J kg-1 K-1. - Physics

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Question

10g of ice at 0℃ absorbs 5460 J of heat energy to melt and change to water at 50℃. Calculate the specific latent heat of fusion of ice. Specific heat capacity of water is 4200 J kg-1  K-1.

Numerical

Solution

Given, Mass of ice = 10g 

Heat energy absorbed (Q) = 5460 J

Specific heat capacity of water = 4200 J kg-1 K-1 = 4.2 J g-1 K-1

Specific latent heat of fusion of ice = ?

(i) Heat energy required to melt the ice at 0° C to water at 0° C (Q1) = m × L

Substituting the values in the formula we get,

Q1 = 10 × L

(ii) Heat energy required to raise temperature from 0° C to 50° C = m × c × rise in temperature

Substituting the values in the formula we get,

Q2 = 10 × 4.2 × 50

Q2 = 2100 J

Then,

Q = Q1 + Q2

∴ 5460 J = 10 × L + 2100 J

∴ 5460 J - 2100 J = 10 × L

∴ 3360 = 10 × L

∴ `3360/10` = L

∴ L = 336 Jg-1

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Chapter 11: Calorimetry - Exercise 11 (B) 3 [Page 281]

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Selina Physics [English] Class 10 ICSE
Chapter 11 Calorimetry
Exercise 11 (B) 3 | Q 1 | Page 281
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