Question

Three different liquids are mixed together in a calorimeter. The masses of the liquids are:
m₁ = 540 g,
m₂ = 400 g,
m₃ = 495 g.
The specific heats are:
c₁ = 215 J/kg°C,
c₂ = 540 J/kg°C,
c₃ = 855 J/kg°C.
The initial temperatures of the liquids are:
T₁ = 22.0 °C,
T₂ = 55.5 °C,
T₃ = 90.0 °C.
What will be the temperature of the mixture in Celsius? (Enter only a number without a unit.)?

Answer 1

Suppose the final temperature is T. (Also assuming T > 55.5 C)

Now Using energy conservation:

Heat gained by m1 and m2 = Heat released by m3

Q1 + Q2 = Q3

m1*C1*dT1 + m2*C2*dT2 = m3*C3*dT3

dT1 = Tf - Ti = T - 22

dT2 = T - 55.5

dT3 = 90 - T

m1 = 0.54 kg & m2 = 0.4 kg & m3 = 0.495 kg

Now using given values:

0.54*215*(T - 22) + 0.4*540*(T - 55.5) = 0.495*855*(90 - T)

Now Solving above equation

T = (0.495*855*90 + 0.54*215*22 + 0.4*540*55.5)/(0.54*215 + 0.4*540 + 0.495*855)

T = 69.7

Please Upvote.

OR you can use this formula directly next time

T = (m1c1T1 + m2c2T2 + m3c3T3)/(m1c1 + m2c2 + m3c3)