A mixture containing 21.3 g of ice (at exactly 0.00 ∘C) and 70.2 g of water (at 52.4 ∘C) is placed in an insulated container.?

Roger the Mole

(333.6 J/g) x (21.3 g) = 7105.68 J to melt the ice (7105.68 J) / (4.184 J/g·°C) / (70.2 g) = 24.2°C lost by the warm water to melt the ice 52.4°C - 24.2°C = 28.2°C the temperature of the warm water after all the ice melts Now you have two bodies of water, 21.3 g at 0°C, and 70.2 g at 28.2°C. When they mix, the final temperature can be found by a weighted average: ((21.3 g x 0°C) + (70.2 g x 28.2°C)) / (21.3 g + 70.2 g) = 21.6°C

david

21.3 X [335 + 4.184(T - 0)] = 4.184(52.4 - T) <<<<< solve for T 7135.5 + 89.12T = 219.24 - 4.184T 93.303T = -6916.26 T = -74.13 <<<< impossible ... all ice does not melt, the Final temp is 0.00 C .... some ice melts lowering the temp of the water to 0.00C .. then the temp. stays at 0