Question

A sample of copper weighing 12.0g is heated to 100.0 degrees celsius and then placed 50.0 mL of water at 26.5 degrees celsius. What will the final temperature of the water be?

Answer #1

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

since density of water is 1 g/ml and volume is 50.0 mL,

m1 = 50.0 g

T1 = 26.5 oC

C1 = 4.184 J/goC

m2 = 12.0 g

T2 = 100.0 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)

12.0*0.385*(100.0-T) = 50.0*4.184*(T-26.5)

4.62*(100.0-T) = 209.2*(T-26.5)

462 - 4.62*T = 209.2*T - 5543.8

T= 28.1 oC

Answer: 28.1 oC

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