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Question
A calorimeter of negligible heat capacity contains 100 cc of water at 40°C. The water cools to 35°C in 5 minutes. The water is now replaced by K-oil of equal volume at 40°C. Find the time taken for the temperature to become 35°C under similar conditions. Specific heat capacities of water and K-oil are 4200 J kg−1 K−1 and 2100 J kg−1 K−1respectively. Density of K-oil = 800 kg m−3.
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Solution
Given:
Volume of water, V = 100 cc = 100×10-3 m3
Change in the temperature of the liquid, ∆θ = 5°C
Time, T = 5 min
For water,
`(msDelta theta)/t =(KA)/l(T_1- T_0 )`
`⇒ (ms)/t = (KA)/l ((T_1-T_0))/(Deltatheta)`
⇒`(100xx10^-3xx1000xx4200)/5 =( KA)/(l) ((313 - T_0))/(Deltatheta)`.........(i)
For K-oil,
`(ms)/t = (KA)/(l)(9(T_1 - T_0))/(Deltatheta)`
⇒ `(ms)/t=(KA)/(l)((T_1 - T_0))/(Deltatheta) `
⇒ `(Vps)/t = (KA)/(l)(T_1 - T_0)/(Deltatheta)`
`⇒ (100xx10^-3xx800xx2100)/(t) = (KA)/(l)((313-T_0))/(Deltatheta)` .............(ii)
From (i) and (ii),
`(100xx10^_3xx800xx2100)/(t) = (100xx10^-5xx1000xx4200)/5`
⇒ t = `(5xx800xx2100)/(1000xx4200) = (2000)/1000`
⇒ t = 2 min
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