5 g of ice at 0^oC is dropped in a beaker containing 20 g of water at 40^oC. What will be the final temperature?

Let final temperature be θ. Heat taken by ice = m₁L + m₁c₁∆θ₁                 = 5 x 80 + 5 x 1(θ-0)                 = 400 + 5θ Heat given by water at 40⁰ C                 = m₂c₂∆θ₂                 = 20 x 1 x (40-θ)                 = 800 -20θ Heat given = Heat taken 800 -20θ   =  400+ 5θ 800-400   =   5θ + 20θ        400      = 25θ               θ     = 400/25                      = 16⁰ C