Q&A - Ask Doubts and Get Answers
Q

Stuck here, help me understand: Steam at 100oC is passed into 20 g of water at 10oC. When water acquires a temperature of 80oC, the mass of water present will be:[Take specific heat of water = 1 cal g-1oC -1and latent heat of steam = 540 cal g-1]

Steam at 100o C is passed into 20 g of water at 10o C. When water acquires a temperature of 80o C, the mass of water present will be:

[Take specific heat of water = 1 cal g-1 oC -1 and latent heat of steam = 540 cal g-1]

  • Option 1)

    24 g

  • Option 2)

    31.5 g

  • Option 3)

    42.5 g

  • Option 4)

    22.5 g

 
Answers (1)
147 Views
V Vakul

As discussed in

Specific Heat -

Amount of heat(Q) required to raise the temperature of unit mass through 1oC.

- wherein

Units-Calories/gm	imes ^{o}C

 

Specific heat of water S_{w} = 1 \:\:cal\:\: g^{-1} \:\: ^{\circ} C^{-1}

Latent heat of steam L_{s}= 540 \:\: cal \:\: g^{-1}

Q_{1} = mL_{s}+mS_{w} \Delta T_{w} = m\times 540+m\times 1\times \left | 100-80 \right |

= 540m+20m = 560 m

Heat gained by 20g of water to change its temperature from 10^{\circ}C to 80^{\circ}C

Q_{2} = m_{w}S_{w}\Delta T_{w} = 20\times 1\times \left | 80-10 \right | = 1400

According to principle of calorimetry Q_{1} = Q_{2}

\therefore 5600w = 1400 or w = 2.5g

total mass of water = = (20+m)g = (20+2.5)g= 22.5 g


Option 1)

24 g

This solution is incorrect

Option 2)

31.5 g

This solution is incorrect

Option 3)

42.5 g

This solution is incorrect

Option 4)

22.5 g

This solution is correct

Exams
Articles
Questions