Question

Determine the specific heat of a 335-gram alloy which is at the boiling point of water, and when put into 300 ml of 10°C H2O reaches equilibrium at 70°C.  How much heat is absorbed by this system? (cwater=1 cal/groC)

Answer 1

t₁(alloy) = 100 ^oC

t₂(alloy) = 70 ^oC

$\Delta$t(alloy) = 100 - 70 = 30 ^oC

Q = $\Delta$t(water) $\times$ m(water)$\times$C(water) = (70 - 10)$\times$ 300$\times$= 18000 (cal)

C (alloy) =$\frac{Q}{(\Delta t(alloy) \times m(alloy))}$ $\frac{Q}{(\Delta t(alloy) \times m(alloy))}$ = $\frac{18000}{(30 \times 335)}$ = 1.791(cal $\times$ g^-1$\times$ ^oC^-1)

Answer: C(alloy) = 1.791 (cal $\times$ g^-1$\times$ ^oC^-1)