2 grams of a substance was dissolved in 75.0 g of water resulting in a temperature decrease from 23.1 ⁰C to 18.7 ⁰C. The specific heat of the solution is 4.18 J/g·⁰C. Calculate the amount of heat associated with the water.

1379.4 Joules

Explanation: The quantity of heat is calculated multiplying the mass of a substance by heat capacity and the change in temperature.

Therefore;

Quantity of heat = Mass * specific heat capacity * Change in temperature

Q = mcΔT

In this case;

The substance dissolved in water gained heat while water lost heat energy.

Thus, Heat gained by the substance = heat lost by water

Heat associated with the water

Mass of water = 75 g

Change in temperature = 4.4°C

Specific heat capacity = 4.18 J/g·⁰C

Heat = mcΔT

= 75 g * 4.18 J/g·⁰C * 4.4 °C

=1379.4 Joules