Question

Calculate the quantity of heat required to convert 1.5kg of ice at 0°C to water at20°C ( latent heat of fusion of ice=3.35×10^5J/ kg, specific heat of water= 4180 J/ kg^-10°C)

ANS::(6.279×10^5J)... the ans provided in the bracket is the correct ans of these que..

Plz solve it tommrow is my exam .. plz

Answer 1

,Total heat = Heat required to convert 1.5 kg of ice to 1.5 kg of water at 0 °C + Heat required to convert 1.5 kg of water at 0 °C to 1.5 kg of water at 20 °C.

Heat=mhfg+mCpΔTHeat=mhfg+mCpΔT

Here, m ( mass of ice) = 1.5 kg

hfg (latent heat of fusion of ice) = 3.35×10^5 J/kg

Cp of water (specific heat) = 4.180 KJ/Kg-K

ΔT(Temperature difference) = 20 °C

Therefore, Heat required = 2 x 3.35×10^5 + 2 x 4.180 x (20 - 0 )

Heat reqd= 6.279×10^5J