Question

Find the equilibrium temperature if 1.5 kg of 150°F ethylene glycol is added to 2 kg of ice/water at 0°C (1.5 kg of liquid water and 0.5 kg of ice) in a 0.5 kg aluminum container which is always the same temperature as the water it contains. The specific heat capacity of ethylene glycol is 2380 J/kg°C.

What is the equilibrium temperature of the contents? Q= ? Celcius

Please show how to go through this problem! Thank you!

Answer 1

Suppose the final temperature is T.

Now Using energy conservation:

Heat gained + Heat released = 0

Q1 + Q2 + Q3 + Q4 = 0

m1*C1*dT1 + (m2 + m3)*C2*dT2 + m3*Lf + m4*C4*dT4 = 0

given,

m1 = mass of ethylene glycol = 1.50 kg

m2 = mass of water = 1.50 kg

m3 = mass of ice = 0.50 kg

m4 = mass of aluminium = 0.50 kg

C1 = Specific heat capacity of ethylene glycol = 2380 J/kg-C

C2 = Specific heat capacity of water = 4186 J/kg-C

C4 = Specific heat capacity of aluminum = 900 J/kg-C

Lf = latent heat of fusion = 3.35*10^5 J/kg

dT1 = Tf - Ti = T - (150 degF) = T - 65.56 (As, C = (F - 32)*5/9)

dT2 = T - 0 = T

dT4 = T - 0 = T

Now using given values:

1.50*2380*(T - 65.56) + (1.5 + 0.5)*4186*(T - 0) + 0.50*3.35*10^5 + 0.50*900*(T - 0) = 0

T = (1.50*2380*65.56 - 0.50*3.35*10^5)/(1.50*2380 + 2.0*4186 + 0.50*900)

T = 5.37 degC

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