An aluminum container of mass 100 g contains 200 g of ice at −20∘C. Heat is added to the system at a rate of 100 cal s−1. Which of the following represents the variation in the temperature of the system as a function of time? Specific heat capacity of ice = 0.5 cal g−1 ∘C−1, specific heat capacity of aluminum = 0.2 cal g−1∘ C−1 and latent heat of fusion of ice = 80 cal g−1.
An aluminum container of mass 100 g contains 200 g of ice at −20∘C. Heat is added to the system at a rate of 100 cal s−1. Which of the following represents the variation in the temperature of the system as a function of time? Specific heat capacity of ice = 0.5 cal g−1 ∘C−1, specific heat capacity of aluminum = 0.2 cal g−1∘ C−1 and latent heat of fusion of ice = 80 cal g−1.
The correct option is B
Total heat supplied to the system in 4 minutes is Q = 100 cal s−1 × 240 s = 2.4 × 104cal.
= (100 g) × (0.2 cal g g−1∘C−1) × (20∘C) + (200 g) × (0.5 cal g−1∘C−1) × (20∘C)
= 400 cal + 2000 cal = 2400 cal
The time taken in this process = 2400100 s = 24 s.
The heat required to melt the ice at 0∘C
= (200 g) (80 cal g−1) = 16000 cal.
The time taken in this process = 16000100 s = 160 s.
If the final temperature is θ, the heat required to take the system to the final temperature is
= (100 g) (0.2 cal g−1∘C−1)θ + (200 g) (1 cal g g−1∘C−1)θ.
Thus,
2.4 θ 104 cal = 2400 cal + 16000 cal + (200 cal∘C−1)θ
or, 5600 cal220 cal∘C−1 = 25.5∘C
The variation in the temperature as a function of time is sketched in diagram
Total heat supplied to the system in 4 minutes is Q = 100 cal s−1 × 240 s = 2.4 × 104cal.
= (100 g) × (0.2 cal g g−1∘C−1) × (20∘C) + (200 g) × (0.5 cal g−1∘C−1) × (20∘C)
= 400 cal + 2000 cal = 2400 cal
The time taken in this process = 2400100 s = 24 s.
The heat required to melt the ice at 0∘C
= (200 g) (80 cal g−1) = 16000 cal.
The time taken in this process = 16000100 s = 160 s.
If the final temperature is θ, the heat required to take the system to the final temperature is
= (100 g) (0.2 cal g−1∘C−1)θ + (200 g) (1 cal g g−1∘C−1)θ.
Thus,
2.4 θ 104 cal = 2400 cal + 16000 cal + (200 cal∘C−1)θ
or, 5600 cal220 cal∘C−1 = 25.5∘C
The variation in the temperature as a function of time is sketched in diagram