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How much heat is released when 233 grams of water at room temperature (25C) is frozen into ice at -15C?

User Aswathy
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Final answer:

To find the heat released when 233 grams of water at 25°C is frozen to ice at -15°C, calculate the heat needed to cool the water to 0°C, the heat released during freezing, and the heat to cool the ice to -15°C.

Step-by-step explanation:

To calculate heat released when 233 grams of water at room temperature (25°C) is frozen into ice at -15°C, we need to consider both cooling the water to 0°C and the latent heat of fusion for converting water to ice, plus the additional cooling to -15°C.

The specific heat capacity of water is 4.18 J/g°C, and the latent heat of fusion for water is approximately 333 J/g. To cool water to 0°C, we use the formula Q = m*c*ΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

To convert water at 0°C to ice at 0°C, we multiply the mass by the latent heat of fusion. Finally, we calculate the heat needed to cool the ice from 0°C to -15°C using the specific heat capacity of ice, which is roughly 2.09 J/g°C.

So the total heat released is the sum of the heat released during these three steps:


  1. Cooling water from 25°C to 0°C: Q1 = m*c*ΔT1

  2. Freezing water at 0°C: Q2 = m*latent heat of fusion

  3. Cooling ice from 0°C to -15°C: Q3 = m*c*ΔT2

If we plug in the values: Q1 = 233g * 4.18 J/g°C * 25°C, Q2 = 233g * 333 J/g, Q3 = 233g * 2.09 J/g°C * 15°C. We can then sum Q1, Q2, and Q3 to find the total heat released.

User Wayne Werner
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