Question

A 1.11 kg piece of aluminum at 78.3 c is put into a glass with 0.210 kg of water at 15.0

Answer 1

M = mass of aluminium = 1.11 kg

$c_{a}$ = specific heat of aluminium = 900

$T_{ai}$ = initial temperature of aluminium = 78.3 c

m = mass of water = 0.210 kg

$c_{w}$ = specific heat of water = 4186

$T_{wi}$ = initial temperature of water = 15 c

T = final equilibrium temperature = ?

using conservation of heat

Heat lost by aluminium = heat gained by water

M $c_{a}$ ($T_{ai}$ - T) = m $c_{w}$ (T - $T_{wi}$ )

(1.11) (900) (78.3 - T) = (0.210) (4186) (T - 15)

T = 48.7 c