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Water has a specific heat of 4.184 J/g°C. If the temperature of 250 g of

water changes from 22.9°C to 14.7°C, how much heat energy was removed
from the water?

User Eatcrayons
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2 Answers

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The amount of heat energy removed from the water can be calculated using the formula:

Q = m x c x ΔT

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

In this case, we have:

m = 250 g
c = 4.184 J/g°C
ΔT = 22.9°C - 14.7°C = 8.2°C (note that the temperature change is negative because the water is losing heat)

Substituting these values into the formula:

Q = 250 g x 4.184 J/g°C x (-8.2°C)
Q = -8,188.4 J

The negative sign indicates that heat energy was removed from the water (as opposed to added). Therefore, 8,188.4 J of heat energy was removed from the water.
User Nkg
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User ArendE
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