Question

An electric heater of power 600 W raises the temperature of 4.0 kg of a liquid from 10.0℃ to 15.0℃ in100 s. Calculate:

1. the heat capacity of 4.0 kg of liquid,
2. the specific heat capacity of the liquid.

Answer 1

Power of heater P = 600 W

t = 100 S

Mass of liquid m = 4.0 kg

Change in temperature of liquid = (15 − 10)°C = 5°C (or 5 K)

Time taken to raise its temperature = 100 s

(i) Heat capacity of liquid C' = `"Q"/(Δ"T")`

∴ C' = `60000/5`

= 12000 JK^-1 or 1.2 × 10⁴ JK^-1

(ii) Specific heat capacity of liquid C = ?

C = `"Q"/("m"Δ"T")`

= `60000/(4 xx 5)`

= 3 × 10³ J Kg^-1°C^-1