1, g of a steam at 100°C melt ........ gm ice at 0°C,? ( Latent heat of ice = 80 ,cal/gm and

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10-1.Thermometry, Thermal Expansion and Calorimetry

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$1\, g$ of a steam at $100°C$ melt ........ $gm$ ice at $0°C\,?$ $($Latent heat of ice $= 80 \,cal/gm$ and latent heat of steam $= 540\, cal/gm)$

A

$1$

B

$2$

C

$4$

D

$8$

Solution

(d) Suppose $m\, gm$ ice melted, then heat required for its melting

$ = mL = m \times 80\,cal$

Heat available with steam for being condensed and then brought to $0°C$

$ = 1 \times 540 + 1 \times 1 \times (100 – 0) = 640\,cal$

$\Rightarrow$ Heat lost $=$ Heat taken

$\Rightarrow$ $640 = m \times 80$ $\Rightarrow$ $m = 8\,gm$

Short trick: You can remember that amount of steam $(m')$ at $100°C$ required to melt m gm ice at $0°C$ is $m' = \frac{m}{8}$.

Here, $m = 8 \times m' = 8 \times 1 = 8\,gm$

Standard 11

Physics

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(IIT-2016)