Steam at 100,^oC is passed into 22,g of water at 20,^oC . mass of water that will be present when wa

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10-1.Thermometry, Thermal Expansion and Calorimetry

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Steam at $100\,^oC$ is passed into $22\,g$ of water at $20\,^oC$. The mass of water that will be present when the water acquires a temperature of $90\,^oC$ (Latent heat of steam is $540\, cal/g$ ) is ............ $\mathrm{g}$

A

$24.8$

B

$24$

C

$36.6$

D

$30$

Solution

Heat loss $=$ Heat gain

$m \times 540+m \times 1 \times(100-90)=22 \times 1(90-20)$

$\mathrm{m}=2.8 \mathrm{\,g}$

net mass of water in mixture $=22+2.8=24.8 \mathrm{\,g}$

Standard 11

Physics

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