How much heat is absorbed by 0.1 kg of ice at -20.0°C to become water at 0.0°C? Consider: Specific heat capacity of ice'= 2060 J/kg.°C, Hf of water = 334000 J/kg

3.75×10⁴ J

Explanation:

We'll begin by calculating the heat required to change the temperature of the ice. This can be obtained as follow:

Mass (M) = 0.1 Kg

Initial Temperature (T₁) = –20 °C

Final temperature (T₂) = 0 °C

Specific heat capacity (C) = 2060 J/kg°C

Heat (Q₁) required to change the temperature of ice =?

Q₁ = MCΔT

Q₁ = 0.1 × 2060 × (0 – –20)

Q₁ = 0.1 × 2060 × (0 + 20)

Q₁ = 206 × 20

Q₁ = 4120 J

Next, we shall determine the heat required to melt the ice. This can be obtained as follow:

Mass (M) = 0.1 Kg

Heat of fusion (Hբ) = 334000 J/kg

Heat (Q₂) required to melt the ice =?

Q₂ = MHբ

Q₂ = 0.1 × 334000

Q₂ = 33400 J

Finally, we shall determine the total heat absorbed. This can be obtained as follow:

Heat (Q₁) = 4120 J

Heat (Q₂) = 33400 J

Total heat (Q) =?

Q = Q₁ + Q₂

Q = 4120 + 33400

Q = 37520

Q = 3.75×10⁴ J