We have half a bucket (6l) of water at 20,^oC . If we want water at 40,^oC , how much steam a

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10-1.Thermometry, Thermal Expansion and Calorimetry

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We have half a bucket $(6l)$ of water at $20\,^oC$ . If we want water at $40\,^oC$, how much steam at $100\,^oC$ should be added to it ?

A

$200\,g$

B

$\frac {200}{9}\,g$

C

$2\,kg$

D

$\frac {200}{3}\,g$

Solution

$6000 \times 1 \times (40-20) = m \times 540 + m \times 1 \times (100 -40)$

$12 \times 104 = m \times 600$

$\Rightarrow m = 2 \times 10^2 = 200\, gm$

Standard 11

Physics

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(JEE MAIN-2019)