Question

A 404−g piece of copper tubing is heated to 89.5°C and placed in an insulated vessel containing 159 g of water at 22.8°C. Assuming no loss of water and a heat capacity for the vessel of 10.0 J/°C, what is the final temperature of the system (c of copper = 0.387 J/g·°C)?

Answer #1

Let us denote water by symbol 1, copper by symbol 2 and calorimeter by symbol 3

m1 = 159.0 g

T1 = 22.8 oC

C1 = 4.184 J/goC

m2 = 404.0 g

T2 = 89.5 oC

C2 = 0.387 J/goC

T = to be calculated

C3 = 10.0 J/g

Let the final temperature be T oC

we have below equation to be used:

heat lost by 2 = heat gained by 1 and 3

m2*C2*(T2-T) = m1*C1*(T-T1) + C3*(T-T1)

404.0*0.387*(89.5-T) = 159.0*4.184*(T-22.8)+10.0*(T-22.8)

156.348*(89.5-T) = 675.256*(T-22.8)

13993.146 - 156.348*T = 675.256*T - 15395.8368

T= 35.3 oC

Answer: 35.3 oC

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