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

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As discussed in

Specific Heat -

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

- wherein

$Units-Calories/gm\times ^{o}C$

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

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