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300 g of water is brought to boiling temperature. The water is then left to cool to room temperature (25°C). The specific heat heat capacity is 4200 J/kg°C. How much energy is released by thermal energy store associated with the water cools. Show working

User ToYonos
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1 Answer

1 vote

Answer:

94.5kJ

Step-by-step explanation:

To calculate the energy released by the thermal energy store associated with the water cooling, we can use the following formula:

Q = mcΔT

where Q is the energy released, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

We first need to calculate the temperature change of the water. The initial temperature of the water is the boiling point of 100°C, and the final temperature is the room temperature of 25°C. Therefore, the temperature change is:

ΔT = (25°C - 100°C) = -75°C

Note that the temperature change is negative because the water is cooling down.

Next, we can substitute the given values into the formula and solve for Q:

Q = (0.3 kg) x (4200 J/kg°C) x (-75°C)

Q = -94500 J

The negative sign indicates that energy is released by the thermal energy store associated with the water cooling. Therefore, the energy released is 94,500 J, or approximately 94.5 kJ.

User Riseres
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