Question

How many grams of ice at -24.9 degrees Celsius can be completely converted to liquid at 5.1 degrees Celsius if the available heat for this process is 4.22x10^3 kJ?

For ice, use a specific heat of 2.01J/(gxC) and enthalpy of fusion of 6.01kJ/mol.

Answer 1

Answer

10420g

Explanation

Heat absorbed by ice (q) = Heat required to raise the temperature of ice from -24.9℃ to 0℃ ( q₁) + Heat required to melt ice at 0℃ (q₂) + Heat required to raise the temperature of melted ice from 0℃ to 5.1℃

q = q₁ + q₂ + q₃

q₁ = m × ∆T × C_s

= m × 24.9℃ × 2.01J/(g × ℃)

= m× 50.049J/g

= m × 0.050049kJ/g

q₂ = m × ∆H_fus

= m × 0.33359J/g

q₃ = m × ∆T × C_s

= m × 5.1℃ × 4.184J/g℃

= m × 21.338J/ g ℃

= m × 0.021338kJ/g

q = 4220kJ

4220kJ =( m × 0.050049kJ/g) + (m × 0.33359kJ/g) + ( m × 0.021338kJ/g )

4220kJ = m × 0.40498kJ/g

m = 10420g