Answer:
T = 74°C
Explanation:
Given Mw = mass of water = 330g, Ma = mass of aluminium = 840g
Cw = 4.2gJ/g°C = specific heat capacity of water and Ca = 0.9J/g°C = specific heat capacity of aluminium
Initial temperature of water = 100°C.
Initial temperature of aluminium = 29°C
When the boiling water is poured into the aluminum pan, heat is exchanged and after a short time the water and aluminum pan both come to thermal equilibrium at a common temperature T.
Heat lost by water equal to the heat gained by aluminium pan.
Mw × Cw×(100 –T) = Ma × Ca × (T–29)
330×4.2×(100– T) = 890×0.9×(T–29)
1386(100 – T) = 801(T –29)
1386/801(100 – T) = T – 29
1.73(100 – T) = T – 29
173 –1.73T = T –29
173+29 = T + 1.73T
202 = 2.73T
T = 202/2.73
T = 74°C
