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 temperature is _______ (specific heat of Cu=420Jkg−1°C−1 and specific heat of water is 4200Jkg−1°C−1
For copper Mass, m1=50g Temperature rise, Δt1=10°C Specific heat,s1=420Jkg−1°C−1 For water Mass, m2=10g Lel rise in temperalure, Δt2= ? Specific heat,s2=4200Jkg−1°C−1 Since, the same amount of heat supplied to copper and water Q1=Q2⇒m1s1Δt1=m2s2Δt2 Substituting the value, we get 50×420×10=10×4200×Δt2 Δt2=5°C