Request a call back

Asked by vinayakumar06361 | 20 Dec, 2023, 11:09: AM

Question # 39

Let m = 1 kg be mass of ice , Cice = 2100 J/kgK is  specific heat of ice , L = 3.34 × 105 J is

latent heat of fusion of ice, Cw is specific heat of water and T is final temperature of water after ice is completely melted.

Heat gain by ice = m Cice ΔT + m × L + m × Cw × T

where ΔT = 10o C is the change in temperature of ice from initial point to melting point.

Heat gained by ice = ( 1 × 2100 × 10 ) + ( 1 × 3.34 × 105 ) + ( 1 × 4186 × T)   J

Heat gained by ice = 3.55 × 105 + ( 4186 × T )   J

Let mw be the mass of water .

Heat lost by water = mw × Cw ( 30 - T ) = 4.4 × 4186 × (30 - T )

By conservation of energy , Heat gained by ice equals heat lost by water.

4.4 × 4186 × (30 - T ) = 3.55 × 105 + ( 4186 × T )

By solving above expresstion for T , we get T = 8.7o C

--------------------------------------------------------------------------------------------------

Question # 40

Let ms be the mass of steam and Lv = 2.260 × 106 J/kgK  be the latent heat of vapouraisation.

Heat lost by steam = ( ms × Lv ) + ( ms ×  Cw ( 100 -90 ) )

Heat lost by steam = ms × [ 2.260 × 106 + 4186 × 10 ] = m× 2.302 × 106 J

Let mw be mass of water.

Heat gain by water = mw × Cw × ΔT = 22 × 10-3 × 4186 × ( 90 - 20)  J  = 6.446 × 103 J

By conservation of energy , heat lost by steam equals heat gain by water .

Hence , we get

m× 2.302 × 106  = 6.446 × 103

ms = 2.8 g

Total mass of water at end = mw + ms = ( 22.0 + 2.8 ) g \ = 24.8 g

Answered by Thiyagarajan K | 20 Dec, 2023, 01:40: PM
NEET neet - Physics
Asked by vinayakumar06361 | 20 Dec, 2023, 11:09: AM