Because all temperatures are below boiling temperature, only sensible heat exchange occurs. The equilibrium temperature is 58 °C. Coffee: initial temperature = 78.7 °C mass = 170 g = 170 x 10⁻³ kg specific heat = 4190 J/(kg-K) temperature drop = 78.7 - 58 = 20.7 K Because the tempeature drops, the coffee loses heat. Heat loss = (170x10⁻³ kg)*(4190 J/(kg-K)*(20.7 K) = 14744.6 J = 14.745 kJ Cream: initial temperature = 7.5 °C Assume that the mass is m kg (not given) specific heat (same as for coffee) temperature rise = 58 - 7.5 = 50.5 K Because the temperature rises, the cream gains heat. Heat gained = (m kg)*(4190 J/(kg-K)*(50.5 K) = 211595m J = 211.595m kJ Cup: initial temperature = 22 °C mass = 115 g = 115 x 10⁻³ kg specific heat = 1091 J/(kg-K) temperature rise = 58 - 22 = 36 K Because of temperature rise, the cup gains heat. Heat gained = (115x10⁻³ kg)*(1091 J/(kg-K))*(36 K) = 4529.2 J = 4.53 kJ Assume no heat is lost to the surroundings. For energy balance (with heat gain as positive and heat loss as negative), obtain -14.475 + 211.595m + 4.53 = 0 m = 0.047 kg = 47 g Answer: 47 g of cream were added, and the cream gained heat.
You pour 170 g hot coffee at 78.7°c and some cold cream at 7.50°c to a 115-g cup that is initially at a temperature of 22.0°c. the cup, coffee, and cream reach an equilibrium temperature of 58.0°c. the material of the cup has a specific heat of 1091 j/(kg · k) and the specific heat of both the coffee and cream is 4190 j/(kg · k). assume that no heat is lost to the surroundings or gained from the surroundings. (a) does the cream lose heat or gain heat?