What is the change in heat energy when 40.0g of water at 25.0oc is raised to 60.0oc?

Given:

40.0g of water

25.0oC is raised to 60.0oC

Required:

Change in heat energy

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

H = (40g) (4.18 J/g-K) (60 – 25)

H = 5,852 J

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