Question

If you were to dump 45.0 g of copper metal (specific heat = 0.092 cal/g°C) at 115°C into 75.0 g of water at 22.3°C, what temperature will the final mixture be?

Answer 1

Heat lost by copper metal = Heat gained by the water

q(Copper) = q(Water)

Let x be the final temperature of the mixture,so,

m(copper).C(copper).deltaT(copper) = m(water).C(water).deltaT(water)

m(copper)=45g

C(copper)=0.092cal/gC

deltaT(copper) = (115-x)C

m(water) = 75g

C(water) = 1cal/gC

deltaT(water) = (x-22.3)C

45g.0.092cal/g°C.(115-x)°C = 75g.1cal/g°C.(x-22.3)°C

Solving for x,

45g*0.092cal/gC / 75g.1cal/gC = (x-22.3)/(115-x)

0.0552= (x-22.3)/(115-x)

6.348 -0.0552x = x-22.3

28.648 = 1.0552x

x = 27.2°C