Calculate the amount of heat gained by the water in cup 2 after adding the hot object(s) to it. Use the data recorded in parts B and C and the formula Q = mCΔT. The specific heat capacity of water is 4.186 J/(g °C).
(part B and C numbers are included)

The amount of heat gained by the water in cup 2 after adding the hot object(s) to it is 2119.121 Joules

Explanation:

As we know

Amount of heat gained

Q = mc (T2-T1)

Here,

mass of water in cup 2 (m) = 79.10 grams

Temperature of water in cup 2 = 16.8 degree Celsius

Specific heat of water (c) = 4.186 J/(g °C)

Final Temperature of water in cup 2  = 23.2 degree Celsius

Substituting the given values, we get -

Q = 79.10 * 4.186 * (23.2 -16.8) = 2119.121 Joules

The amount of heat gained by the water in cup 2 after adding the hot object(s) to it is 2119.121 Joules

the volume increases