How much heat in kcal is required to change 0.5 kg of ice, originally at - 10 0 * C into steam at 110 C?Constants needed in the problemLatent heat of fusion=79.7 kcal/kg Specific heat of ice=0.5 kcal/kg/K ; Latent heat of vaporization ation = 539 kcal/kg ; Specific heat of water 1.0 kcal/kg/K Specific heat of ieam=0.480 kcal/kg

Q = 364.25 kcal

Explanation:

In this question, we will have to calculate the heat absorptions for different steps of temperature rise and phase change. And then we will ad them to calculate total heat absorbed.

1. RISE IN TEMPERATURE OF ICE:

First, the temperature of ice will be increased from - 10°C to 0 °C. Heat absorbed during this process will be given as:

Q₁ = mC₁ΔT₁

where,

Q₁ = Heat absorbed while increasing temperature of ice = ?

m = mass of ice = 0.5 kg

C₁ = specific heat of ice = 0.5 kcal/kg k

ΔT₁ = change in temperature of ice = 0 - (-10) = 10 k

Therefore,

Q₁ = (0.5 kg)(0.5 kcal/kg.k)(10)

Q₁ = 2.5 kcal

2. MELTING OF ICE:

Now, the melting of ice will occur at 0°C and the heat absorbed during this process will be:

Q₂ = m(Latent Heat of Fusion of Ice)

where,

Q₂ = heat Absorbed during melting of ice = ?

Therefore,

Q₂ = (0.5 kg)(79.7 kcal/kg)

Q₂ = 39.85 kcal

3. RISE IN TEMPERATURE OF WATER:

Now, the temperature of water will be increased from 0°C to 100 °C. Heat absorbed during this process will be given as:

Q₃ = mC₃ΔT₃

where,

Q₃ = Heat absorbed while increasing temperature of water = ?

m = mass of water = 0.5 kg

C₃ = specific heat of water = 1 kcal/kg k

ΔT₃ = change in temperature of ice = 100 - 0 = 100 k

Therefore,

Q₃ = (0.5 kg)(1 kcal/kg.k)(100 k)

Q₃ = 50 kcal

4. VAPORIZATION OF WATER:

Now, the vaporization of water will occur at 100°C and the heat absorbed during this process will be:

Q₄ = m(Latent Heat of Vaporization of Water)

where,

Q₄ = heat Absorbed during vaporization of water = ?

Therefore,

Q₄ = (0.5 kg)(539 kcal/kg)

Q₄ = 269.5 kcal

5. RISE IN TEMPERATURE OF STEAM:

Now, the temperature of steam will be increased from 100°C to 110 °C. Heat absorbed during this process will be given as:

Q₅ = mC₅ΔT₅

where,

Q₅ = Heat absorbed while increasing temperature of steam = ?

m = mass of steam = 0.5 kg

C₅ = specific heat of steam = 0.48 kcal/kg k

ΔT₅ = change in temperature of ice = 110 - 100 = 10 k

Therefore,

Q₅ = (0.5 kg)(0.48 kcal/kg.k)(10 k)

Q₅ = 2.4 kcal

Hence, the total heat absorbed to change 0.5 kg of ice at - 10°C into steam at 110°C will be:

Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Q = 2.5 kcal + 39.85 kcal + 50 kcal + 269.5 kcal + 2.4 kcal

Q = 364.25 kcal