Question

An electric heater raises the temperature of 5000g of a given liquid. The electric heater is rated 1 KW power.

To raise the temperature of the liquid from 25⁰C to 31⁰C, a heater required 120 seconds. Calculate i) the heat

capacity of the liquid, ii) The specific heat capacity of the liquid.​

Answer 1

Answer:

Explanation:

Time t = 2 minutes = 2 x 60 = 120 s

Energy supplied by the heater = power x time

Energy used in raising the temperature of liquid from 25oC to 31oC

Answer 2

Given info : An electric heater raises the temperature of 5000g of a given liquid. The electric heater is rated 1 KW power. To raise the temperature of the liquid from 25⁰C to 31⁰C, a heater required 120 seconds.

To find : Calculate

1. the heat capacity of the liquid
2. the specific heat capacity of the liquid

solution : mass of liquid, m = 5000 g = 5 kg

power , P = 1 kW = 1000 W

time, t = 120 seconds

so energy = power × time = 1000 × 120 = 1,20,000 J

we know, heat capacity = amount of heat/raising in temperature

= 120000/(31°C - 25°C)

= 120000/6

= 20000 J/°C

= 2 × 10⁴ J/°C

Therefore the heat capacity of the liquid is 2 × 10⁴ J/°C

now specific heat capacity = heat capacity/mass

= 2 × 10⁴ J/°C/5 kg

= 0.4 × 10⁴ J/kg /°C

= 4 kJ/kg/°C

Therefore the specific heat capacity of the Liquid is 4 kJ/kg/°C