1300 J of heat energy is supplied to raise the temperature of 0.5 kg of lead from 20oC to 40oC. Calculate the specific heat capacity of lead. - Noon Academy

1300 J of heat energy is supplied to raise the temperature of 0.5 kg of lead from 20^oC to 40^oC. Calculate the specific heat capacity of lead.

Calorimetry | Class 10 | ICSE | Physics | Selina

1300 J of heat energy is supplied to raise the temperature of 0.5 kg of lead from 20^oC to 40^oC. Calculate the specific heat capacity of lead.

Solution:

According to the question,

Heat energy supplied is 1300 J

and the mass of lead is 0.5 kg

Also the change in temperature is

(40 – 20)⁰ C = 20⁰ C

Expression for specific heat capacity of lead is ==> C = △Q / m△T

Upon substituting values, we get

C = 1300 / 0.5 × 20 = 130 J kg^-1 K^-1

Therefore, the specific heat capacity of lead is 130 J kg^-1 K^-1.

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