## Chemistry 12th Edition

(a) We know that $P.E=mgh$ We plug in the known values to obtain: $P.E=0.001\times 9.81\times 51$ $P.E=0.5J$ (b) As $K.E=\frac{1}{2}mv^2=0.5J$ We plug in the known values to obtain: $\frac{1}{2}\times 0.001\times v^2=0.5$ $v^2=1000$ $v=32\frac{m}{s}$ (c) We know that $q=ms\Delta T$ We plug in the known values to obtain: $0.5J=(0.001\times 1000)\times 4.184\times \Delta T$ $\Delta T=0.12C^{\circ}$