What mass of steam initially at 120oC is needed to warm 200g of water in a glass container from 20.0 oC to 50.0 oC

Complete question:

What mass of steam initially at 120 ⁰C is needed to warm 200g of water in a 100 g glass container from 20.0 oC to 50.0 ⁰C

the initial mass of the steam is 10.82 g

Explanation:

Given;

mass of water, m₁ = 200 g

mass of the glass, m₂ = 100 g

temperature of the steam = 120 ⁰C

initial temperature of the water, 20⁰ C

final temperature of the water, = 50⁰ C

let the mass of the steam = m

specific heat capacity of water c = 1 cal/g ⁰ C

specific heat capacity of glass c₂ = 0.2 cal/g ⁰ C

laten heat of vaporization of steam L = 540 cal/g

Apply principle of conservation energy;

Heat given off by the steam = Heat absorbed by water + heat absorbed by glass

m(1) (120 - 100)  +  m(540)  + m(1) (100 - 50) = [200(1)  +  100(0.2)] (50 - 20)

20m + 540m + 50m = 6600

610 m = 6600

m = 6600 / 610

m = 10.82 g

Therefore, the initial mass of the steam is 10.82 g

- lead + phosphoric acid = hydrogen + lead(ii) phosphate