explain?
A hot iron horseshoe (mass = 0.40 kg) is dropped into 1.6 L of water in a 0.3 kg iron pot initially at 20°C. If the final equilibrium temperature is 25°C, estimate the initial temperature of the hot horseshoe.

Question

explain?
A hot iron horseshoe (mass = 0.40 kg) is dropped into 1.6 L of water in a 0.3 kg iron pot initially at 20°C. If the final equilibrium temperature is 25°C, estimate the initial temperature of the hot horseshoe.

jennychan12 · Answer

mshoecshoe ΔTshoe = (mpotcpot + mwatercwater) ΔTpot ;
(0.40 kg)(450 J/kg · C°)(T – 25°C) =
[(0.30 kg)(450 J/kg · C°) + (1.60 L)(1.00 kg/L)(4186 J/kg · C°)](25°C – 20°C),
which gives T = 215°C.
explain?
i got 62.95 degrees celsius