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 electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.