A piece of copper at 125c

Question

A piece of copper at 125 °C is placed in a styrofoam coffee cup containing 29.0 g water at 25.0 °C. The final temperature of both substances at thermal equilibrium is 35.0 °C. Assuming that no heat is lost to the cup or the surroundings, what is the mass of the copper piece? The specific heat capacity of copper Cs = 0.385 J/g°C and water = 4.18 J/g°C

Answer 1

Ali,

Use the formula q = mC∆T to find the heat gained by the water when placing hot copper in it.

q = heat

m = mass of water = 29.0 g

C = specific heat of water = 4.18 J/gº

∆T = change in temperature = 35.0º - 25º = 10º

q = (29.0 g)(4.18 J/gº)(10º) = 1212 J

Since the water gained 1212 J of heat, the copper must have LOST 1212 J of heat. We can use the same formula now to find the mass of copper:

q = mC∆T

q = heat = 1212 J

m = mass of copper = ?

C = specific heat of copper = 0.385 J/gº

∆T = change in temperature of copper = 125º - 25º = 100º

1212 J = (m)(0.385 J/gº)(100º)

m = 1212 J / 38.5 J/g)

m = 31.5 g (3 sig. figs.)

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Done as one operation, where heat gained by water must = heat lost by copper:

heat gained by water = heat lost by copper

(29.0 g)(4.18 J/gº)(10º) = (m)(0.385 J/gº)(100º)

1212 = 38.5 m

m = 31.5 g (3 sig. figs.)