Question

Calculate the amount of energy when water at 72 degrees c freezes completely at 0 degrees c

Answer 1

Answer:

The amount of energy needed when water at 72 degrees c freezes completely at 0 degrees c is $6.37\times10^4$ Joules

Explanation:

$Q = m \times c \times \Delta T$

where

$\Delta T$ = Final T - Initial T

$Q_{1}=100 g \times 4.184 \frac{J}{g^{\circ} \mathrm{C}} \times\left(72^{\circ} \mathrm{C}-0^{\circ} \mathrm{C}\right)$

$Q_1$ =30125J

Q is the heat energy in Joules  

c is the specific heat capacity (for water 1.0  cal/(g℃))  or 4.184  J/(g℃)

m is the mass of water

mass of water is assumed as 100 g (since not mentioned)

$Q_2$  is the heat energy required for the phase change

$Q_2$ =mass × heat of fusion

$\\ Q_{2}=100 g \times 336 \frac{J}{g} \\\\ Q_{2}=33600 J$

Total heat = $Q_1 + Q_2$

Total Heat = 30123J + 33600J

= 63725 J

= $6.37\times10^4$ Joules is the answer