How many joules are required to change 1kg of ice at-10°c to water at 15°c?

Explanation:

There are three heat transfers in this process:

Total heat = warming the ice + melting the ice + warming the water  

       q        =           q₁               +           q₂             +              q₃

       q        =       mC₁ΔT₁          +        mΔHfus       +          mC₂ΔT₂

Let's calculate these heat transfers separately.

Data:

      m = 1 kg = 1000 g

     C₁ = 2.108 J·°C⁻¹g⁻¹

     C₂ = 4.184 J·°C⁻¹g⁻¹

ΔHfus = 333 J/g

Tmax =   15 °C

 m.p. =    0 °C

Tmin = -10 °C

Calculations:

ΔT₁ = 0 – (-10) = 10 °C

q₁ = 1000 × 2.108 × 10 = 21 080J = 21.08 kJ

q₂ = 1000 × 333 = 333 000 J =333 kJ

ΔT₂ = 15 - 0 = 15 °C

 q₃ = 1000 × 4.184 × 15 = 62 760 J = 62.76 kJ

q = 21.08 + 333 + 62.76 = 417 kJ

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