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?

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.