Question

What is the final equilibrium temperature when13gofmilkat23◦Cisaddedto132gof coffee at 60◦C? Assume the specific heats of milk and coffee are the same as that of water, and neglect the specific heat of the container. Answer in units of ◦C.

Answer 1

Given data:
mass of milk = 13gr
at T = 23 degree celsius
mass of coffee = 132 grm
at T = 60 degree celsius

the heat gained by the milk will equal the heat lost by the coffee; the heat gain/loss of a substance is

Q = m c delta T

m=mass of sample
c=specific heat
delta T = change in temp

therefore, we have:

mass milk x c x (T-23) = mass coffee x c x (60-T)

the common terms of c cancel, leaving us with

13g x (T-23) = 132g x (60-T)

solve for T: T=56.68 degree celsius