Numerical Problem No. 8.8 (Thermal Properties of Matter) Physics SSC - I (Class 9)

Find the quantity of heat needed to melt 100g of ice at -10°C into the water at 10°C. Ans(39900 J) 

(Note: Specific heat of ice is 2100 Jkg⁻¹K⁻¹, the specific heat of water is 4200 Jkg⁻¹K⁻¹, and Latent heat of fusion of ice is 336000 Jkg⁻¹).

Given:

Mass of ice  = m = 100 g = 0.1 kg

Initial Temperature = T₁ = -10°C = (-10 + 273) K = 263 K

Final Temperature = T₂ = 10°C = (10 + 273) K = 283 K

Change in Temperature = ΔT = T₂  - T₁ = 283 K - 263 K = 20 K

Specific heat of Ice  = c_w = 2100 J kg⁻¹ K⁻¹

Specific heat of water  = cᵢ = 4200 J kg⁻¹ K⁻¹

Latent heat of fusion of ice  = H_f  = 336,000 J kg⁻¹

To Find:

Heat  = Q  = ?

Solution: 

The quantity of heat Q needed to melt 0.1 Kg of ice at
-10°C into the water at 10°C will be equal to the sum of 

(1)  Heat transformation Q₁  during latent heat of fusion, 

(2) the Heat Q₂ needed for the change of ice to water from -10°C to 0°C  

and 

(3) Heat Q₃ needed to increase the temperature from 0°C to 10°C. 

Q  = Q₁ + Q₂ + Q₃    ----eqn (1)

(1)  Heat transformation Q₁  during latent heat of fusion, 

We have the formula for Latent heat of fusion is

Q₁ = m H_f

By putting values 

Q₁ = 0.1 Kg ｘ 336,000 J kg⁻¹

Q₁ = 33,600 J

(2) The Heat Q₂ needed for the change of ice to water from -10°C to 0°C  

Here using the formula for heat capacity

Q₂  = mcΔT

By Putting values (from temperature range -10°C to 0°C the ΔT = 10 K)

Q₂  = 0.1 Kg ｘ 2100 J kg⁻¹K⁻¹ ｘ 10 K 

Q₂ = 2,100 J

(3) Heat Q₃ needed to increase the temperature from 0°C to 10°C.   

Here using the formula for heat capacity

Q₃  = mcΔT

By Putting values (from temperature range 0°C to 10°C the ΔT = 10 K)

Q₃  = 0.1 Kg ｘ 4200 J kg⁻¹ K⁻¹ ｘ 10 K  

Q₃ = 4,200 J

Hence using Eqn (1)

Q  = Q₁ + Q₂ + Q₃

Q  = 33,600 J + 2,100 J + 4,200 J

Q  = 39,900 J

Thus the total heat needed to melt the required ice is 39,900 J

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