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What is the change in temperature in a 128g sample of water if it absorbs 2808J of heat energy at a temperature of 2°C?

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Given:

128g sample of water

2808J of heat energy

Required:

Change in temperature

Solution:

This can be solved through the equation H = mCpT where H is the heat, m is the mass, Cp is the specific heat and T is the change in temperature.

The specific heat of the water is 4.18 J/g-K

Plugging in the values into the equation

H = mCpT

2808J = (128g) (4.18 J/g-K) T

T = 5.25 K

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