50g of copper is heated to increase its temperature by 10∘C. If the same quantity of heat is given to 10g of water, the rise in its temperature is (specific heat of copper = 420J/kg/∘C, specific heat of water = 4200J/kg/∘C)
5°C
Given
Mass of copper = 50g
Mass of water = 10g
specific heat of copper = 420J/kg/∘C
specific heat of water = 4200J/kg/∘C
Change in temperature = ΔT
Same amount of heat is supplied to copper and water; so Q1 = Q2
mcccΔTc = mωcωΔTω
⇒ (ΔT)ω = mcccΔTcmωcω = 50 × 10−3 × 420 × 1010 × 10−3 × 4200 = 5∘C