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Grade 10Thermal Physics

a refrigerator converts 100g of water at 200 C to ice at -100 C in 73.5 min. calculate the average rate of heat extraction in watt. the specific heat capacity of water is 4.2jg-1K-1, specic latent heat of ice is 336Jg-1K-1 and specific haet capacity of ice is 2.1 Jg-1K-1

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11 Years agoGrade 10
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1 Answer

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Askiitians Tutor Team

ApprovedApproved Tutor Answer1 Year ago

To find the average rate of heat extraction by the refrigerator, we need to calculate the total heat removed during the entire process of converting water at 20 °C to ice at -10 °C. This involves three main steps: cooling the water from 20 °C to 0 °C, freezing the water at 0 °C to ice, and then cooling the ice from 0 °C to -10 °C. After calculating the total heat removed, we can then determine the average rate of heat extraction in watts.

Step 1: Cooling Water from 20 °C to 0 °C

First, we need to calculate the heat removed when cooling the water. The formula for heat transfer is:

Q = mcΔT

  • m = mass of water = 100 g
  • c = specific heat capacity of water = 4.2 J/g·K
  • ΔT = change in temperature = (0 - 20) °C = -20 K

Now, substituting the values:

Q₁ = 100 g × 4.2 J/g·K × (-20 K) = -8400 J

Step 2: Freezing Water at 0 °C to Ice

Next, we calculate the heat removed during the phase change from water to ice. The formula for latent heat is:

Q = mL

  • L = specific latent heat of ice = 336 J/g

Substituting the values:

Q₂ = 100 g × 336 J/g = 33600 J

Step 3: Cooling Ice from 0 °C to -10 °C

Finally, we calculate the heat removed when cooling the ice. Again, we use the heat transfer formula:

  • c = specific heat capacity of ice = 2.1 J/g·K
  • ΔT = change in temperature = (-10 - 0) °C = -10 K

Substituting the values:

Q₃ = 100 g × 2.1 J/g·K × (-10 K) = -2100 J

Total Heat Removed

Now, we can find the total heat extracted by summing up all the heat values:

Total Q = Q₁ + Q₂ + Q₃

Total Q = -8400 J + (-33600 J) + (-2100 J) = -44100 J

Calculating Average Rate of Heat Extraction

To find the average rate of heat extraction, we divide the total heat removed by the time taken. The time is given as 73.5 minutes, which we need to convert to seconds:

Time = 73.5 min × 60 s/min = 4410 s

Now, we can calculate the average rate:

Average Power (P) = Total Q / Time

P = -44100 J / 4410 s ≈ -10 W

Final Result

The average rate of heat extraction by the refrigerator is approximately 10 watts. The negative sign indicates that heat is being removed from the system, which is expected in refrigeration processes.