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Question
A calorimeter contains 50 g of water at 50°C. The temperature falls to 45°C in 10 minutes. When the calorimeter contains 100 g of water at 50°C, it takes 18 minutes for the temperature to become 45°C. Find the water equivalent of the calorimeter.
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Solution
Let water equivalent to calorimeter be w.
Change in temperature = 5°C
Specific heat of water = 4200 J/Kg °C
Rate of flow of heat is given by
q = Energy per unit time = `(msDeltaT)/t`
Case 1:
`q_1 = ((w +50 xx 10^-3 )xx 4200xx5)/10`
Case 2
`q^2 = ((w+100 xx10^-3)xx 4200xx5)/18`
From calorimeter theory, these two rates of flow of heat should be equal to each other.
⇒ q1 = q2
`((w + 50 xx 10^-3)xx 4200xx5)/10 = ((w + 100xx10^-3)xx4200xx5)/18`
`rArr 18(W + 50 xx 10^-3) = 10(W + 100 xx10^-3)`
⇒ W = 12.5 ×10-3 kg
⇒ W = 12.5 g
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