Question

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

Answer 1

`\boxed{\text{417 kJ}}`

Step-by-step 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

You must add
`\boxed{\textbf{417 kJ}}`.