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)
And=122.5g
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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