50g of copper is heated to increase its temperature by 10∘C. If the same quantity of heat is given to 10g of water, then what would be the rise in its temperature?
(specific heat of copper =420J/kg/∘C, specific heat of water =4200J/kg/∘C)
5∘C
Given,
Mass of copper, mc=50g
Rise in temperature, ΔTc=10∘C
Mass of water, mω=10g
Same amount of heat is supplied to copper and water, so
mcccΔTc = mωcωΔTω
where, m, c and ΔT is the mass, specific heat and change in temperature of copper and water respectively.
or, (ΔT)ω=mcccΔTcmωcω
=50×10−3×420×1010×10−3×4200
=5∘C