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
Answer:
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
[tex]mc\Delta T_1 + mL + mc\Delta T_2 = m_1c\Delta T_3 + m_2c_2\Delta T_3\\\\mc\Delta T_1 + mL + mc\Delta T_2 = [m_1c + m_2c_2]\Delta T_3[/tex]
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
