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  • Help me understand! Ice at -200C is added to 50g of water at 400C. When the temperature of the mixture reaches 00C, it is found that 20g of ice is still unmelted. The amount of ice added to the water was close to : (Specific heat of water = 4.2J/g/

Ice at -200C is added to 50g of water at 400C. When the temperature of the mixture reaches 00C, it is found that 20g of ice is still unmelted. The amount of ice added to the water was close to :

(Specific heat of water = 4.2J/g/ 0C

Specific heat of ice = 2.1J/g/ 0C

Heat of fusion of watre at 00C = 334J/g )

 

  • Option 1)

    100 g

  • Option 2)

    50g

  • Option 3)

    40 g

  • Option 4)

    60g

Answers (1)

best_answer

 

Mixture of two substance -

\theta_{mix}=\frac{m_{1}c_{1}\theta_{1}+m_{2}c_{2}\theta_{2}}{m_{1}c_{1}+m_{2}c_{2}}

- wherein

\theta_{min}= Temperaure of mixture at equilibrium. 

 

 

Mixture of two substance -

\theta_{mix}=\frac{m_{1}c_{1}\theta_{1}+m_{2}c_{2}\theta_{2}}{m_{1}c_{1}+m_{2}c_{2}}

- wherein

\theta_{min}= Temperaure of mixture at equilibrium. 

Say amount of ice = m gm

heat taken by ice = heat given by water

20\times 2.1\times m+(m-20)334

=50\times 4.2\times 4.0

\Rightarrow m=40.1\approx 4.0

 

 

 


Option 1)

100 g

Option 2)

50g

Option 3)

40 g

Option 4)

60g

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