Answer:
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.
<u>1. RISE IN TEMPERATURE OF ICE:</u>
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
<u>2. MELTING OF ICE:</u>
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
<u>3. RISE IN TEMPERATURE OF WATER:</u>
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
<u>4. VAPORIZATION OF WATER:</u>
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
<u>5. RISE IN TEMPERATURE OF STEAM:</u>
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
<u>Q = 364.25 kcal</u>