Question

Caculate the temperature change if 3456 J of thermal energy was transferred into 3.0 kg of lead? The specific heat of lead is 128 J/kgo C.

Answer 1

9°C

Explanation:

quantity of heat energy absorbed is expressed as;

Q = mcΔt

m is the mass

c is the specific heat capacity

Δt is the change in temperature

Substitute the given values

3456 = 3 * 128 Δt

3456 = 384Δt

Δt = 3456/384

Δt = 9°C

Hence the temperature change is 9°C