Asked by Anita Iwu

steam at 100 degree C is passed into a container of negligible heat capacity containing 20 g of ice and 100 g of water at 0 degree C, until the ice is completely melted. Determine the total mass of water in the container. ( specific latent heat=2•3×10^3, specific latent heat of ice= 3•4×10^2, specific heat capacity of water = 4•2)

Answered by bobpursley
Well, one has to know the initial temperature of the ice.
Answered by Doyin
Answer the question
Answered by Cash
Answer
Answered by Stella Shammah
Pls somebody should answer.Plss.
Answered by Success
Pls,I really need the answer
Answered by Eno monday
And=122.5g
Answered by Paul Ogboye
Q1 = Q2
Where Q1 - heat of melting ice,
Q2 - heat of condensation of vapor
m1 - mass of ice = 20g
m2 - mass of condensation of vapor(steam) = ?
L1 - specific latent heat of ice = 340
L2 - specific latent heat of steam = 2300
Q1 = m1 * L1
Q1 = 20 * 340 = 6800J
since Q1 = Q2
Q2 = m2 * L2
6800 = m2 *2300
m2 = 6800/2300
m2 = 2.9 g
Therefore, mass of water in the container = 100+20+2.9= 122.9 g
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