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10-1.Thermometry, Thermal Expansion and Calorimetry
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One kilogram of ice at $0\,^oC$ is mixed with one kilogram of water at $80\,^oC$. Temperature of mixture is ........ $^oC$ (specific heat of water $= 4200\,J\,kg^{-1}\,K^{-1}$ and latent heat of ice $= 336\,KJ \,kg^{-1}$ )
A
$40$
B
$60$
C
$0$
D
$50$
Solution
$\begin{array}{ll}{1000 \mathrm{gm}} & {1000 \mathrm{gm}} \\ {\mathrm{ice}} & {\text { water }} \\ {0^{\circ} \mathrm{C}} & {80^{\circ} \mathrm{C}}\end{array}$
heat energy required for melting entire ice $\mathrm{Q}=\mathrm{mL}=1000 \times 80=80,000 \mathrm{cal}$
Heat given by water
$\mathrm{Q}=\mathrm{ms} \Delta \mathrm{T}=1000 \times 1 \times 80=80,000 \mathrm{cal}$
temp. of mixture $=0^{\circ} \mathrm{C}$
Standard 11
Physics
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