Answer:
The amount of energy needed when water at 72 degrees c freezes completely at 0 degrees c is [tex]6.37\times10^4[/tex] Joules
Explanation:
[tex]Q = m \times c \times \Delta T[/tex]
where
[tex]\Delta T[/tex] = Final T - Initial T
[tex]$Q_{1}=100 g \times 4.184 \frac{J}{g^{\circ} \mathrm{C}} \times\left(72^{\circ} \mathrm{C}-0^{\circ} \mathrm{C}\right)$[/tex]
[tex]Q_1[/tex] =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)
[tex]Q_2[/tex] is the heat energy required for the phase change
[tex]Q_2[/tex] =mass × heat of fusion
[tex]\\$Q_{2}=100 g \times 336 \frac{J}{g}$\\\\$Q_{2}=33600 J$[/tex]
Total heat = [tex]Q_1 + Q_2[/tex]
Total Heat = 30123J + 33600J
= 63725 J
= [tex]6.37\times10^4[/tex] Joules is the answer
