25.6 kg of liquid 1 comes in thermal contact with 74.4 kg of liquid 2. If liquid

Question

25.6 kg of liquid 1 comes in thermal contact with 74.4 kg of liquid 2. If liquid 1 has an initial temperature of 372K, and a specific heat of 4.12 kJ/kg K, and liquid 2 has ayn initial temperature of 287 K, and a specific heat of 3.34 kJ/kg K, then what is the final temperature at thermal equilibrium? Assume the system is thermally isolated. 25.6 kg of liquid 1 comes in thermal contact with 74.4 kg of liquid 2. If liquid 1 has an initial temperature of 372K, and a specific heat of 4.12 kJ/kg K, and liquid 2 has ayn initial temperature of 287 K, and a specific heat of 3.34 kJ/kg K, then what is the final temperature at thermal equilibrium? Assume the system is thermally isolated.

Explanation / Answer

m1 = 25.6 Kg
T1 = 372 K
C1 = 4.12 KJ/Kg.K
m2 = 74.4 Kg
T2 = 287 K
C2 = 3.34 KJ/Kg.K
T = to be calculated

Let the final temperature be T oC
we have below equation to be used:
heat gained by 2 = heat lost by 1
m2*C2*(T-T2) = m1*C1*(T1-T)
74.4*3.34*(T-287) = 25.6*4.12*(372-T)
248.496*(T-287) = 105.472*(372-T)
248.496*T -71318.352 = 39235.584 - 105.472*T
T= 312 K
Answer: 312 K

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