#### 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^{−1}respectively. Density of K-oil = 800 kg m^{−3}.

#### Solution

Given:

Volume of water, *V* = 100 cc = 100×10^{-3} m^{3}

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