Question

How much steam at 100°C, will just melt down 3200 g of ice at - 10C?
Given : Specific heat capacity of ice = 2100 x J kg C
Specific heat capacity of water = 4200 J kg- c
Specific latent heat of steam - 2260 x 10^3 J kg
Specific latent heat of ice 336x 10^3 J Kg-1​

Answer 1

Answer:

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Answer 2

Answer:

Explanation:

M- 3200g - 3.2kg

T- -10°c

S ice- 2100 j kg

S water- 4200 j kg

L steam - 2260 × 10^3 j kg

L ice - 336 × 10^3 j kg

Q1 - need to convert -10°c to 0°c

M. S ice. ∆t

3.2× 2100× 10= 67200 j

Q2- convert ice to water

M. S water ∆ t

3.2× 4200×10= 134400 j

Q3- to bring the water to boil ( 0 to 100°c)

M. L steam. ∆t

3.2 × 2260 × 10^3 = 7232000 j

Q4 (100°c to steam)

M. L ice ∆t

3.2 × 336× 10^3 = 1075200 j

Q = add Q1+Q2+Q3+Q4 = 8508800 j