Question

If 45 g of silver at 270oC are mixed with 300 g of water at 20oC, find the final temperature of the
system.

Answer 1

Assume no heat loss to the surroundings and no phase changes. Let Tf be the final temperature.

Heat lost by silver = heat gained by water:
mAg cAg (TAg − Tf) = mw cw (Tf − Tw)

Solve for Tf:
Tf = (mAg cAg TAg + mw cw Tw) / (mAg cAg + mw cw)

Take cAg ≈ 0.24 J/g·°C and cw ≈ 4.18 J/g·°C.
mAg = 45 g, TAg = 270 °C; mw = 300 g, Tw = 20 °C.

Tf = (45·0.24·270 + 300·4.18·20) / (45·0.24 + 300·4.18)
= (2,916 + 25,080) / (10.8 + 1,254)
= 27,996 / 1,264.8 ≈ 22.1 °C.

Final temperature ≈ 22.1 °C.