Question

The specific heat of water is quite high compared to the specific heat of lead, so the energy released by a simply of heated lead to its surroundings will be only around 4% of the amount of heat by an equal mass of water under identical conditions.
a) calculate the change in temperature that results from the addition of 2500J of heat to a 25g sample of water.
b) calculate the change in temperature that results from the addition of 2500J of heat energy to a 25g sample of lead. The specific heat of lead is 0.126J/(g K).
c) explain the difference in the temperature changes, found in parts (a) and (b), in terms of the specific heat capacities of lead and water.

Answer 1

a) Q= mc∆T.

∆T= Q/mc= 2500/25x 4.18=23.923445 °C.

b)

∆T= Q/mc= 2500 /25 x 0.126 = 793.650794.

c) ∆T2/∆T1= 793.650794/23.923445 = 33.1746032.

C1/C2=4.18/0.126 =33.1746032

Means how many times less heat capacity of the substance, the higher the temperature