Question

Suppose I pour 13.0 mL of creamer into 220 mL of coffee. The creamer is at 33.7 deg F, and the coffee is at 208 deg F.

What is the final temperature when the system reaches equilibrium, in deg F?

Assume coffee and creamer have the same density and thermal properties of water.

Answer 1

let rho is the density and c is the specific heat of the water.

let T is the final temperature

heat gained by creamer = heat lost by coffee

m_creamer*C*(T - 33.7) = m_coffee*C*(208 - T)

13*C*(T - 33.7) = 220*C*(208 - T)

13*T - 13*33.7 = 220*208 - 220*T

T*(13 + 220) = 220*208 + 13*33.7

T = (220*208 + 13*33.7)/(13 + 220)

= 198 F