Question

A man buys a 12 oz (355 mL) cup of coffee 176 degrees F. He pours 0.25 oz (7.5 mL) milk at 41 degrees F into the coffee. If the milk has specific heat cal = 4010 J/kg degree C and coffee has the same specific heat as water, what is the final temperature of this mixture?

Answer 1

mass of coffee, m = density * Volume

m = 1.3*355 = 461.5 g

(density of milk = 1.03 kg/m^3)

mass of milk, m' = 1.03*7.5 = 7.725 g

Let the final temperature = T

Heat lost by coffee,

Q1 = m*c(T1 - T)

Heat gained by milk,

Q2 = m'*c*(T - T2)

Here, Q1 = Q2

m*c(T1 - T) = m'*c*(T - T2)

0.4615*4186*(176 - T) = 7.725*10^(-3)*4010*(T - 41)

T = 178.2 deg