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ICSE Class 10 Physics 2019 Paper

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Question : 67 of 74
Marks: +1, -0
104g104 \text{g} of water at 30∘C30^{\circ} \text{C} is taken in a calorimeter made of copper of mass 42g42 \text{g} . When a certain mass of ice at 0∘C0^{\circ} \text{C} is added to it, the final steady temperature of the mixture after the ice has melted, was found to be 10∘C10^{\circ}\text{C}. Find the mass of ice added.
[Specific heat capacity of water =4.2Jg−1∘C−1=4.2 \mathrm{Jg}^{-1} {}^{\circ} \mathrm{C}^{-1} ;
Specific latent heat of fusion of ice =336Jg−1;=336 \mathrm{Jg}^{-1};
Specific heat capacity of copper =0.4Jg−1=0.4 \mathrm{Jg}^{-1} ∘C−1]{}^{\circ} \mathrm{C}^{-1}]
Solution:  
Let the mass of ice added be == ' mm ' grams
According to the principle of calorimetry,
Heat gained by ice to melt and then convert to water at 10∘C10^{\circ} \text{C}
= Heat lost by water + Heat lost by calorimetry.
mL+mCWΔT  =mW×CW×ΔTw+mCm L + m \mathrm{C}_{\mathrm{W}} \Delta T \; = m_{\mathrm{W}} \times \mathrm{C}_{\mathrm{W}} \times \Delta T_{\mathrm{w}} + m_{\mathrm{C}} ×CC×ΔTC\times \mathrm{C}_{\mathrm{C}} \times \Delta T_{\mathrm{C}}
m×336+m  ×4.2×(10−0)m \times 336 + m \; \times 4.2 \times (10-0) =104×4.2×(30−10)+42×0.4×(30−10)=104 \times 4.2 \times (30-10) + 42 \times 0.4 \times (30-10)
336m+42m  =8736+336336 m + 42 m \; = 8736 + 336
378m  =9072378 m \; = 9072
m  =  9072378=24gm \; = \; \frac{9072}{378} = 24 \text{g}
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