Question

How much heat energy must be added to 52kg Of water at 68°F to raise the temperature to 212°F? The specific heat capacity for water is 4.186×10 to the third power J/kg times degrees Celsius

Answer 1

Answer:

The amount of energy added to rise the temperature Q = 17413.76 KJ

Explanation:

Mass of water = 52 kg

Initial temperature $T_{1}$ = 68 °F = 20° c

Final temperature $T_{2}$ = 212 °F = 100° c

Specific heat of water  $C = 4.186 \frac{KJ}{kg c}$

Now heat transfer Q = m × C × ( $T_{2}$  - $T_{1}$ )

⇒ Q = 52 × 4.186 × ( 100 - 20 )

⇒ Q = 17413.76 KJ

This is the amount of energy added to rise the temperature.