Question

Calculate the specific heat of lead, given that 125 g of the metal absorbs 1600 J when temp is changed from 12⁰C to 112⁰C.

Answer 1

THE SPECIFIC HEAT OF LEAD HERE IS 0.128J/(g°C)

We know that,

q=Cp×m×ΔT-----1

Know quantities:

Heat=q=1600J

Mass=m=125g

ΔT=112°C-12°C=100°

The above equation 1 can be rearranged to solve for specific heat:

Cp= q/(m×ΔT)

Cp=1600/(125*100)

Cp=16/125

Cp=0.128 J/(g°C)

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