NEET Class neet Answered

Question # 39

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

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

Heat gain by ice = m C_ice ΔT + m × L + m × C_w × T

where ΔT = 10^o 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 m_w be the mass of water .

Heat lost by water = m_w × C_w ( 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 = 8.7^o C

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Question # 40

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

Heat lost by steam = ( m_s × L_v ) + ( m_s ×  C_w ( 100 -90 ) )

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

Let m_w be mass of water.

Heat gain by water = m_w × C_w × 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_s  × 2.302 × 10⁶  = 6.446 × 10³ 

m_s = 2.8 g

Total mass of water at end = m_w + m_s = ( 22.0 + 2.8 ) g \ = 24.8 g