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Physics Thermodynamics Mix Subjective Type
Published on: September 12, 2026

A certain amount of ice is supplied heat at a constant rate for 7 minutes. For the first 1 minute, the temperature rises uniformly with time, then it remains constant for the next 4 minute and again rises at a uniform rate for the last 2 minutes. Explain physically these observations and calculate the final temperature. Latent heat of ice = 336 × 10 3 J kg –1 and c water = 4200 J kg –1 K –1 .

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Text Solution

Verified by Experts
The correct answer is:
0 °C
Step 1: Understanding the process
When heat is supplied to ice, it undergoes different processes depending on its temperature. In this scenario:
  • For the first 1 minute: The ice is still at a temperature below 0 °C. The heat supplied increases the temperature of the ice.
  • For the next 4 minutes: The temperature remains constant at 0 °C. During this period, the heat supplied is used for melting the ice into water (phase change). This is where the latent heat is absorbed.
  • For the last 2 minutes: After the ice has completely melted, the resulting water is now being supplied heat, and its temperature begins to rise.
Step 2: Calculate heat supplied
Let the rate of heat supply be Q watts. Hence, the total heat supplied in 7 minutes (420 seconds) is:
$$ Q_{total} = Q imes 420 $$
Step 3: Energy use in first 1 minute (heating ice)
Assuming the mass of ice is m kg, the heat used to raise the temperature of ice from -T to 0 °C is:
$$ Q_1 = m imes c_{ice} imes T $$
Where $c_{ice} = c_{water}$ (approximate) = 4200 J/kg/K and $T = 1°C$ (from -1°C to 0°C).
Step 4: Energy use during phase change (melting ice)
Heat used during the phase change (0 °C ice to 0 °C water):
$$ Q_2 = m imes L_{ice} = m imes 336 imes 10^3 $$
Where $L_{ice}$ is the latent heat of fusion.
Step 5: Energy use in last 2 minutes (heating water)
Assuming all the ice has melted into water, the heat used to raise the temperature of the water (0 °C) to final temperature (T_f) is:
$$ Q_3 = m imes c_{water} imes (T_f - 0) = m imes 4200 imes T_f $$
Step 6: Equating heat supplied and heat used
Putting everything together,
$$ Q imes 420 = Q_1 + Q_2 + Q_3 $$
$$ Q imes 420 = m imes 4200 imes 1 + m imes 336 imes 10^3 + m imes 4200 imes T_f $$
Assuming 1 kg of ice (m = 1):
$$ Q imes 420 = 4200 + 336000 + 4200 imes T_f $$
Step 7: Solve for T_f
Cancelling the units:
$$ Q imes 420 = 340200 + 4200 imes T_f $$
Step 8: Analyze the problem
Since there is no more heat available to raise the temperature above 0 °C as all heat went into the phase change, the final temperature still remains at 0 °C.
Conclusion:
Therefore, the final temperature is 0 °C.

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