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ICSE Class 10 Physics 2016 Paper

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Question : 39 of 73
Marks: +1, -0
A copper vessel of mass 100 g100\,\mathrm{g} contains 150 g150\,\mathrm{g} of water at 50∘ C50^{\circ}\,\mathrm{C}. How much ice is needed to cool it to 5∘ C5^{\circ}\,\mathrm{C} ?
Given :
Specific heat capacity of copper =0.4 Jg−1 ∘C−1=0.4\,\mathrm{Jg}^{-1}\,{}^{\circ}\mathrm{C}^{-1}.
Specific heat capacity of water =4.2 J g−1 C−1=4.2\,\mathrm{J}\,\mathrm{g}^{-1}\,\mathrm{C}^{-1}.
Specific latent heat of fusion of ice =336=336  Jg−1\,\mathrm{Jg}^{-1}
Solution:  
Heat given by water to reach 5∘ C+5^{\circ}\,\mathrm{C}+ Heat given by copper vessel to reach 5∘ C=5^{\circ}\,\mathrm{C}= Heat taken by ice to melt at 0∘ C0^{\circ}\,\mathrm{C} + Heat taken by melted ice to reach 5∘ C5^{\circ}\,\mathrm{C}
mcΔt+mcΔt=mL+mcΔtm c \Delta t+m c \Delta t=m L+m c \Delta t
where Δt=\Delta t= change in temperature
[150×4.2×(50∘−5∘)]+[100×0.4×[150 \times 4.2 \times (50^{\circ}-5^{\circ}) ] +[100 \times 0.4 \times (50∘−5∘)](50^{\circ}-5^{\circ}) ]
=(m×336)+[m×4.2×(5−0)]=(m \times 336)+[m \times 4.2 \times(5-0)] (150×4.2×45)+(100×0.4×45)(150 \times 4.2 \times 45)+(100 \times 0.4 \times 45)
=(m×336)+(m×4.2×5)=(m \times 336)+(m \times 4.2 \times 5)
28350+1800  =336m+21m28350+1800\;=336 m+21 m
357m  =30150357 m\;=30150
m  =  30150357m\;=\;\frac{30150}{357}
  =84.45 gm  of ice.   \;=84.45\ \mathrm{gm}\ \text{ of ice. }\;
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