Question

Five (5.0) liters of water in a plastic liquid dispenser have warmed to 15 degress Cel. In an effort to cool the water 1.0 kg of ice at a temperature of -10 degree Cel. is added to the dispenser. Once the ice melts what is the final temperature of the commbined water? (Ignore the temperature change of the dispenser itself.) (Hints: Don't forget to calculate the heat needded to bring the ice to 0 degreses Cel., and once that ice melts you must treat its 0 degree Cel. water separately from the 15 degree Cel wter.)

Answer 1

Mass of water, m = 5 kg Temp. of water, T_w = 15 degree c Mass of ice, M_i = 1 kg Temp. of ice, T_i = - 10 degree c Specific heat of water s = 4186 J/kg c Latent heat of ice L = 3.33 times 10^5 J/kg Total heat that can be given by water reach 0 degree c H_1 = (M) (S) (15) H_1 = 313250 J Heat needed by ice (11) to melt H_2 (M_i) (L) H_2 = 333000 J So H_2 greaterthan H_1 { full ice cant be melt } Let x melts then (x) (L) = H_1 x = 313250/3.33 times 10^5 x = 0.94 kg Final temp. is 0 degree c