Question

Calculate how much steam from water boiling at 100 C will just melt 50 gm of wax at 20 C. (Melting point of wax=55 C, sp. heat capacity of wax=0.7cal/g C, sp. latent heat of fusion of wax=35cal/g)​

Answer 1

To melt 50 g of wax at 20 °C, approximately 5.52 g of steam is required when considering the specific heat capacity of wax (0.7 cal/g °C) and the specific latent heat of fusion of wax (35 cal/g).

Step-by-step explanation:

To calculate how much steam from water boiling at 100 °C would be required to just melt 50 g of wax at 20 °C with a melting point of 55 °C, the specific heat capacity of wax being 0.7 cal/g °C, and the specific latent heat of fusion of wax being 35 cal/g, we need to find the total heat absorbed by the wax to undergo phase change from solid to liquid.

First, we calculate the heat required to raise the temperature of the wax from 20 °C to its melting point, which is 55 °C:

Q1 = mass × specific heat capacity × change in temperature
Q1 = 50 g × 0.7 cal/g °C × (55 °C - 20 °C)
Q1 = 50 × 0.7 × 35
Q1 = 1225 cal

Next, we compute the heat required for the phase change from solid to liquid at the melting point:

Q2 = mass × latent heat of fusion
Q2 = 50 g × 35 cal/g
Q2 = 1750 cal

The total heat (Q_total) the wax absorbs is the sum of Q1 and Q2:

Q_total = Q1 + Q2
Q_total = 1225 cal + 1750 cal
Q_total = 2975 cal

This is the amount of energy that must be provided by the steam, which condenses to water at 100 °C. Assuming steam undergoes a phase change at constant temperature (100 °C), we will use the latent heat of vaporization of water, which is 539 cal/g, to find the mass of steam required:

mass of steam (m_steam) = Q_total / latent heat of vaporization
m_steam = 2975 cal / 539 cal/g
m_steam ≈ 5.52 g

Therefore, approximately 5.52 g of steam is required to just melt 50 g of wax.

Answer 2

m₁ = 5.507 x 10⁻³ kg = 5.507 g

Step-by-step explanation:

The heat released by the steam must be absorbed by the melting wax, according to the law of conservation of energy:

Heat Released by Steam = Heat Absorbed by Wax

Latent Heat of Fusion of Steam = Heat Absorbed by Wax + Latent Heat of Fusion of Wax

m₁H₁ = m₂CΔT₂ + m₂H₂

where,

m₁ = mass of water = ?

H₁ = Latent Heat of Fusion of Water = 2260 KJ/kg = 2.26 x 10⁶ J/kg

m₂ = mass of wax = 50 g = 0.05 kg

C = specific heat capacity of wax = (0.7 cal/g°C)(1000 g/1 kg)(4.184 J/1 Cal)

C = 2928.8 J/kg.°C

ΔT₂ = Change in Temperature of Wax = 55°C - 20°C = 35°C

H₂ = Latent Heat of Fusion of Wax = (35 cal/g)(1000 g/1 kg)(4.184 J/1 Cal)

H₂ = 146440 J/kg

Therefore,

m₁(2.26 x 10⁶ J/kg) = (0.05 kg)(2928.8 J/kg.°C)(35°C) + (0.05 kg)(146440 J/kg)

m₁ = (5125.4 J + 7322 J)/2.26 x 10⁶ J/kg

m₁ = 5.507 x 10⁻³ kg = 5.507 g