Two hundred fifty grams of water at 80'C is poured into a Styrofoam cup of negligible heat capacity containing 180g of water at 10'C. After an additional 300g of water is added to the cap, the mixture comes to the equilibrium temperature of 30'C. What was the temperature of the additional 300g of water?
-
heat lost by warm water = heat gained by cooler water
the heat lost by the 250 g of water is
Q = 0.25kg x 4186J/kg/C x 50C = 52325J
the heat gained by the 180g of water is
Q = 0.18kg x 4186J/kg/C x 20C = 15069J
since more energy is lost by the warm water than is gained by the cool water, the additional 300g of water must have been colder than 30C and thus gained water
the heat gained by the additional 300q is 52325J - 15069J = 37256J
and we have
37256J = 0.3kg x 4186J/kg/C x delta T
delta T = 29.7 C, so the temp of the 300 g of water was 30-29.7 = 0.3C
the heat lost by the 250 g of water is
Q = 0.25kg x 4186J/kg/C x 50C = 52325J
the heat gained by the 180g of water is
Q = 0.18kg x 4186J/kg/C x 20C = 15069J
since more energy is lost by the warm water than is gained by the cool water, the additional 300g of water must have been colder than 30C and thus gained water
the heat gained by the additional 300q is 52325J - 15069J = 37256J
and we have
37256J = 0.3kg x 4186J/kg/C x delta T
delta T = 29.7 C, so the temp of the 300 g of water was 30-29.7 = 0.3C