Question

A hot lump of 27.5 g of copper at an initial temperature of 54.7 °C is placed in 50.0 mL of H2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the copper and water given that the specific heat of copper is 0.385 J/(g·°C)? Assume no heat is lost to surroundings.

Answer #1

Let us denote water by symbol 1 and copper by symbol 2

m1 = 50.0 g

T1 = 25.0 oC

C1 = 4.184 J/goC

m2 = 27.5 g

T2 = 54.7 oC

C2 = 0.385 J/goC

T = to be calculated

Let the final temperature be T oC

we have below equation to be used:

heat lost by 2 = heat gained by 1

m2*C2*(T2-T) = m1*C1*(T-T1)

27.5*0.385*(54.7-T) = 50.0*4.184*(T-25.0)

10.5875*(54.7-T) = 209.2*(T-25.0)

579.1363 - 10.5875*T = 209.2*T - 5230

T= 26.4 oC

Answer: 26.4 oC

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