2 kg ice at -20^o C is mixed with 5 kg water at 20^o C . Then final amount of water in m

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10-1.Thermometry, Thermal Expansion and Calorimetry

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$2\ kg$ ice at $-20^o\ C$ is mixed with $5\ kg$ water at $20^o\ C$. Then final amount of water in the mixture would be ; Given specific heat of ice $= 0.5\ cal/g^o\ C$, specific heat of water $= 1\ cal/g^o\ C$, Latent heat of fusion of ice $= 80\ cal/g$ ........ $kg$

A

$6$

B

$5$

C

$4$

D

$2$

Solution

$\mathrm{Q}_{1}=2000 \times 0.5 \times 20+2000 \times 80$

$=180000$

$\mathrm{Q}_{2}=5000 \times 1 \times 20$

$=100000$

$\mathrm{T}=0^{\circ} \mathrm{C}$

$\mathrm{Q}=\mathrm{mL}$

$80000=m \times 80$

$\mathrm{m}=1000 \mathrm{gm}$

$=1 \mathrm{kg}$

Ice $=2-1=1 \mathrm{kg}$

$\mathrm{H}_{2} \mathrm{O}=5+1=6 \mathrm{kg}$

Standard 11

Physics

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