S.3 Phy Latent Heat

(1)
Total energy released by steam to become all water
= 2.26 * 10^6 * 1
= 2.26 * 10^6 J

Total energy required to melt all ice
= 3.34 * 10^5 * 1
= 3.34 * 10^5 J

Let m be the mass of steam required to melt the ice.
2.26 * 10^6 * m = 3.34 * 10^5
m = 0.147788

So, all ice will melt into water. The final outcome will be a mixture of steam and water.

Total energy released by remaining steam to become all water
= 2.26 * 10^6 * (1-m)
= 1925999.12 J

Total energy required to raised the (m+1)kg water from 0 to 100 degree Celsius
= 1.147788 * 4200 * 100
= 482070.96

Assume the final mixture will be 100 degree Celsius of steam and water. Let m2 be the total mass of steam which has transformed to water.

2.26 * 10^6 * m2 = 3.34 * 10^5 + (1+m2) * 4200 * 100
m=0.40978 kg

So. the final mixture will be steam and water at 100 degree Celsius.

(2)
Let m be the mass of the steam introduced.
2.26 * 10^6 * m = 40 * 10^-3 * 3.34 * 10^5
m = 0.0059115 kg

(1)
Heat absorbed when 1 kg of ice completely melts then changes to 1kg of 100oC water
= mLf + mcDT
= (1) x (3.34 x 105) + (1) x (4200) x (100 - 0)
= (3.34 x 105) + (4.2 x 105)
= 7.54 x 105 J

Heat released when 1 kg of steam completely changes to 100oC water
= mLv
= (1) x (2.26 x 106)
= 2.26 x 106 J > 7.54 x 105 J

Therefore, the steam would NOT completely condense, and the mixture contains 100oC steam and 100oC water.

The final temperature = 100oC



(2)
Let the mass of steam introduced be y kg.
Then, the mass of melted ice = (0.04 - y) kg.

Heat released when y kg of steam condenses
= mLv
= (y) x (2.26 x 106) J

Heat absorbed when (0.04 - y) kg of ice melts
= (0.04 - y) x (3.34 x 106) J

(y) x (2.26 x 106) = (0.04 - y) x (3.34 x 105)
(2.26 x 106)y = (1.336 104) - (3.34 x 105)y
(2.594 x 106)y = (1.336 104)
y = 0.00515

Mass of steam = 0.00515 kg = 5.15 g
=