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How much is lost or gained when 1250 grams of of water heats from 32 degrees celsius to 100 degrees celsius? The specific heat capacity of water is 4.184 J/g Deg C.

1 Answer

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To calculate the amount of heat lost or gained by a substance when its temperature changes, we can use the formula:

Q = m * c * ΔT

where Q is the amount of heat lost or gained (in Joules), m is the mass of the substance (in grams), c is the specific heat capacity of the substance (in J/g°C), and ΔT is the change in temperature (in °C).

In this case, we want to find the amount of heat gained by 1250 grams of water when its temperature increases from 32°C to 100°C. The specific heat capacity of water is 4.184 J/g°C.

First, we need to calculate the change in temperature:

ΔT = 100°C - 32°C = 68°C

Now we can use the formula to calculate the amount of heat gained:

Q = 1250 g * 4.184 J/g°C * 68°C

Q = 361,768 J or approximately 362 kJ

Therefore, the amount of heat gained by 1250 grams of water when its temperature increases from 32°C to 100°C is approximately 362 kJ.
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User SKulibin
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