Answer to Question #174377 in Chemistry for Ava Brooke

Question #174377

A 24.93 g sample of ice is heated from -16 ^oC to its melting point of 0 ^oC and then melted completely. Next it is raised to its boiling point of 100 ^oC and boiled completely. Finally the sample is heated to 113 ^oC. How much energy was absorbed by the sample? (Answer in kJ pay close attention to units!)

S_ice=2.1 J/g^oC Delta H_f = 6.02 kJ/mol

S_steam=1.7 J/g^oC Delta H_v = 2.26 kJ/g

Expert's answer

Solution:

The molar mass of H₂O is 18.0 g mol^-1.

Hence,

(24.93 g H₂O) × (1 mol H₂O / 18.0 g H₂O) = 1.385 mol H₂O

Moles of H₂O = 1.385 mol

This problem can be summarized thusly:

1) Ice is heated from -16°C to 0°C.

The heat absorbed is calculated by using the specific heat of ice (S_ice= 2.1 J/g°C) and the equation:

q = m × S × ΔT

q₁ = m × S_ice × (T_f - T_i)

q₁ = 24.93 g × 2.1 J/g°C × (0 - (-16))°C = 837.65 J = 0.83765 kJ

q₁ = 0.83765 kJ

2) Ice is melted at 0°C. The heat absorbed is calculated by multiplying the moles of ice by the molar heat of fusion (ΔH_f).

q₂ = ΔH_f × n = 6.02 kJ/mol × 1.385 mol = 8.3377 kJ

q₂ = 8.3377 kJ

3) Water at 0°C is heated to 100°C. The heat absorbed is calculated by using the specific heat of water (S_water= 4.2 J/g°C) and the equation: q = m × S × ΔT

q₃ = m × S_water × (T_f - T_i)

q₃ = 24.93 g × 4.2 J/g°C × (100 - 0)°C = 10470.6 J = 10.4706 kJ

q₃ = 10.4706 kJ

4) Water is vaporized to steam at 100°C. The heat absorbed is calculated by multiplying the grams of water by the heat of vaporization (ΔH_v) in kJ/g.

q₄ = ΔH_v × m = 2.26 kJ/g × 24.93 g = 56.3418 kJ

q₄ = 56.3418 kJ

5) Steam is heated from 100°C to 113°C. The heat absorbed is calculated by using the specific heat of steam (S_steam= 1.7 J/g°C) and the equation: q = m × S × ΔT

q₅ = m × S_steam × (T_f - T_i)

q₅ = 24.93 g × 1.7 J/g°C × (113 - 100)°C = 550.853 J = 0.55095 kJ

q₅ = 0.55095 kJ

q = q₁+ q₂ + q₃ + q₄ + q₅

q = 0.83765 kJ + 8.3377 kJ + 10.4706 kJ + 56.3418 kJ + 0.55095 kJ = 76.5387 kJ = 76.54 kJ