Question

(a) Calculate the amount of heat released in both Joules () and calories (cal) when 15.5 g of liquid water at 22.5 C cools to ice at - 10.0 degrees * C .

Answer 1

The heat absorbed or released by a substance when its temperature changes depend on its heat capacity (Cp). The Cp of the water is 1 cal/g°C. The heat released can be found by the following equation:

`Q=mCp\Delta T`

Where,

Q is the heat released or absorbed

m is the mas oof water, 15.5g

Cp is the heat capacity of water, 1cal/g°C

dT is the change in temperature

`\begin{gathered} \Delta T=T_2-T_1=-10.0\degree C-22.5\degree C \\ \Delta T=-32.5\degree C \end{gathered}`

We replace the known values

`Q=15.5g*1(cal)/(g\degree C)*(-32.5\degree C)`
`\begin{gathered} Q=15.5g*1(cal)/(g\degree C)*(-32.5\degree C) \\ Q=-503.75cal \end{gathered}`

In Joules the heat released will be:

`Q=-503.75cal*(4.1868J)/(1cal)=-2019J`

the heat has a negative value, this is because the water released heat.

Answer:

The amount of heat released is 504 calories or 2019 joules