Question

Calculate the amount of heat released when $5.0g$ of water at $20{}^{\circ }C$is changed into ice at $0{}^{\circ }C$.
(Specific heat capacity of water $=4.2J/g{}^{\circ }C$,
Specific latent heat of fusion of ice $=336J/g$)

Answer 1

Step 1:Given data

mass of water ($m$) = $5.0g$

change in temperature of water ($\Delta T$)$=20{}^{\circ }C$

Specific heat capacity of water ($s$)$=4.2J/g{}^{\circ }C$,

Latent heat of fusion of ice ($L$)$=336J/g$

finding the total heat released ($Q_T$)

Step 2: Calculating the amount of heat released when the water cools from $20{}^{\circ }C$ to $0{}^{\circ }C$:

Let $Q_1$ be the heat released when the water cools from $20{}^{\circ }C$ to $0{}^{\circ }C$.

Therefore,$Q_1$ is given as:

$Q_1=ms\Delta T$

Substituting the given values above:

$\therefore Q_1=5\times 4.2\times 20$

$\therefore Q_1=420J$

Step 3:Calculating the amount of heat released in phase change of water to ice at $0{}^{\circ }C$

Let ‘$Q_2$’ be the amount of heat released during the phase change.

Therefore $Q_2$is given as:

$Q_2=mL$

Substituting the given values above:

$\therefore Q_2=5\times 336$

$\therefore Q_2=1680J$

Step 4:Calculating the total heat released:

Let ‘$Q$’ be the total heat released during the given conversion.

Thus, $Q$ is given as:

$Q=Q_1+Q_2$

$\therefore Q=420+1680$

Thus, $Q=2100J$

Hence, $2100J$ of heat is released when $5.0g$ of water at $20{}^{\circ }C$is changed into ice at $0{}^{\circ }C$.