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Two containers are bought in thermal contact. Container 1 held a liquid with specific heat 4.03 kJ/kg K at a temperature of 306 K. Container 2 held another liquid with specific heat 3.51 kJ/kg K at a temperature of 173 K. If these 2 containers held the same mass of liquid, what is the temperature once thermal equilibrium is reached?

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Answer:

The temperature once the thermal equilibrium is reached is 244.1 Kelvin

Step-by-step explanation:

Step 1: Data given

Container 1 has a liquid with specific heat = 4.03 kJ/kg*K

this liquid is at a temperature of 306 K

Container 2 has a liquid with specific heat = 3.51 kJ/kg*K

this liquid is at a temperature of 173 K

Step 2: Calculate the temperature at equilibrium

ΔU1 + ΔU2 = 0

Q1 +Q2 = 0

m1*c1*(T-T1) = -m2*c2*(T-T2)

⇒ with m1 = m2

⇒ with c1 = the specific heat of liquid 1 = 4.03 kJ/kg*K

⇒ with T1 = 306 K

⇒ with m2 = m1

⇒ with c2 = the specific heat of liquid 2

⇒ with T1 = 173 K

Since m1 = m2 we can write this formula as followed:

c1*(T-T1) = -c2*(T-T2)

4.03 kJ/kg*K *(T- 306K) = - 3.51 kJ/kg*K *( T-173K)

4.03 T - 1233.18 = -3.51 T + 607.23

7.54 T = 1840.41

T = 244.1 K

The temperature once the thermal equilibrium is reached is 244.1 Kelvin

User Andrii Muzalevskyi
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