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)

m₁ = 5.507 x 10⁻³ kg = 5.507 g

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

area of focus

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer