Question

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

Answer 1

The amount of energy needed when water at 72 degrees c freezes completely at 0 degrees c is
`6.37*10^4` Joules

Step-by-step explanation:

`Q = m * c * \Delta T`

where

`\Delta T` = Final T - Initial T

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

`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 * 336 (J)/(g)$\\\\$Q_(2)=33600 J$`

Total heat =
`Q_1 + Q_2`

Total Heat = 30123J + 33600J

= 63725 J

=
`6.37*10^4` Joules is the answer