Question

Question 25 4 pts What mass of ice (ing) can be melted if 5.02 kJ of thermal energy are added at the freezing point? Use molar mass = 18.02 g/mol Question 26 O pts EXTRA CREDIT: How much energy (in kJ) is required to convert 40.00 g of H2O from solid at -5.000 °C to gas at 105.0 °C?

Answer 1

Q. 25 Answer:

Heat of fusion of water = 334 J/g

We know the relation among heat energy, mass and heat of fusion: q = m × ΔH_f

Where, q = heat energy, m = mass, and ΔH_f = heat of fusion

Given, q = 5.02 kJ = 5020 J, ΔH_f = 334 J/g, and m = ?

m = q / ΔH_f = 5020/334 g = 15.02 g

∴ The mass of melted ice = 15.02 g

Q. 26 Answer:

We need to consider-

specific heat of ice (c) = 2.108 J ºC⁻¹g⁻¹,

specific heat of water (c) = 4.184 J ºC⁻¹g⁻¹

specific heat of water vapor (c) = 2.108 J ºC⁻¹g⁻¹,

ΔH_vap = 2260 J g^-1 and ΔH_fus = 334 J g^-1

For this calculation, we have to consider five steps of heat change:

q₁ = heat required to warm the ice from -5.000ºC to 0.00 °C.

q₂ = heat required to melt the ice to water at 0.00°C.

q₃ = heat required to warm the water from 0.00°C to 100.00°C.

q₄ = heat required to vaporize the water to vapor at 100°C.

q₅ = heat required to warm the vapour to 105.0°C.

q₁ = m×c×ΔT = 40.00 g × 2.108 J ºC⁻¹g⁻¹ × {0.00ºC – (-5.00ºC)} = 421.6 J

q₂ = m×ΔH_fus = 40.00 g × 334 J g^-1 = 13360 J

q₃ = m×c×ΔT = 40.00 g × 4.184 J ºC⁻¹g⁻¹ × (100.00ºC – 0.00ºC) = 16736 J

q₄ = m×ΔH_vap = 40.00 g × 2260 J g^-1 = 90400 J

q₅ = m×c×ΔT = 40.00 g × 2.108 J ºC⁻¹g⁻¹ × (105.00ºC – 100.00ºC) = 421.6 J

Therefore, the required heat

q _total = q₁ + q₂ + q₃ + q₄ + q₅

= (421.6 + 13360 + 16736 + 90400 + 421.6) J = 121339.2 J = 121.33 kJ