Question

A hot iron horseshoe (mass = 0.45 kg ), just forged, is dropped into 1.40 L of water in a 0.50 kgiron pot initially at 21.0 ∘C.

If the final equilibrium temperature is 31.0 ∘C, estimate the initial temperature of the hot horseshoe.

Express your answer using two significant figures.

T =    ∘C

Answer 1

Q(pot) = m x c x delta t*C

Q(pot) = 0.50 x 450 x (31-21) ...( specific heat of iron = 450 J/kg.ºC)

Q(pot) = 2250 J

Q(water) = m x c x delta t*C

Q(water) = 1.40 x 4186 x (31 - 21) .....(specific heat of iron = 4186 J/kg.ºC)

Q(water) = 58604 J

Let the initial temperature of the horseshoe was t*C

Q(horseshoe) loss = Q(pot) gain + Q(water) gain

m x c x delta t*C = 2250 + 58604

0.40 x 450 x (t-31) = 60854

180t - 5580 = 66434

t = 65354/180 = 369 C