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A 145.87 g piece of ice is removed from a refrigerator at -11 C. It is placed in a bowl in which it melts and eventually warms to room temperature at 25 C. Calculate the amount of heat (kJ) gained by the sample.

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Final answer:

The piece of ice gains a total of 52.08 kJ of heat as it melts and warms to room temperature.

Step-by-step explanation:

To calculate the amount of heat gained by the piece of ice, we need to consider two steps:

Step 1: Bring the ice from -11°C to 0°C. The heat gained in this step can be calculated using the specific heat capacity of ice:

q₁ = mass × specific heat capacity × change in temperature

q₁ = 145.87 g × 2.09 J/g°C × (0 - (-11))°C

q₁ = 3296.13 J

= 3.30 kJ

Step 2: Melt the ice at 0°C. The heat gained in this step can be calculated using the heat of fusion:

q₂ = mass × heat of fusion

q₂ = 145.87 g × 334 J/g = 48784.58 J

= 48.78 kJ

The total amount of heat gained by the ice is the sum of q₁ and q₂:

Total heat gained = q₁ + q₂

= 3.30 kJ + 48.78 kJ

= 52.08 kJ

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