Question

You pour 220 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 59.0°C. The material of the cup has a specific heat of 0.2604 kcal/(kg · °C) and the specific heat of both the coffee and cream is 1.00 kcal/(kg · C). If no heat is lost to the surroundings or gained from the surroundings, how much cream did you add?

Answer 1

here,

mass of cofee , mk = 0.22 kg

mass of cream is mc

mass of cup , mcup = 0.115 kg

initial temprature of cup , T1 = 22 degree C

initial temprature of coffee , T2 = 78.7 degree C

initial temprature of cream , T3 = 7.5 degree C

final temprature , Tf = 59 degree C

heat lost by cofee = heat gained by cream + heat gained by cup

mk * Cc * ( T2 - Tf) = mc * Cc * ( Tf - T3 ) + mcup * Ccup *( Tf - T1)

0.22 * 1* 19.7 = mc * 1 * 51.5 + 0.115 * 0.2604 * 37

mc = 0.063 kg = 63 g

the mass of the cream is 63 g