Water and turpentine oil (specific heat less than that of water) are both heated to same temperature. Equal amounts of both are then placed in identical calorimeters and then left in air :

A and B will represent cooling curves of water and oil respectively

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B and A will represent cooling curves of water and oil respectively

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Their cooling curves will be identical

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None of the above

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Solution

The correct option is C A and B will represent cooling curves of water and oil respectively
We know that,
Rate of cooling $\propto \frac{1}{C (specific heat)}$
Since $C_{o i l} < C_{w a t e r}$ (given)
$∴$ Rate of cooling of water < Rate of cooling of oil.
It is clear that at a particular time after starting of cooling, temperature of oil will be less than that of water.
So, graph A represents the cooling curve of water and graph B represents the cooling curve of oil.

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. Water and turpentine oil (specific heat less than that of water) are both heated to same temperature. Equal amounts of these placed in identical calorimeters are, then left in air. ‘T’ is temperature of body and t is time taken to cool, then

Water and turpentine oil (specific heat less than that of water) are both heated to same temperature. Equal amounts of these placed in identical calorimeters are then left in air

Q. The rate of cooling of 100g of water at

77^{o} C

is 80cal/min. If the water is replaced by an equal volume of turpentine oil in the same calorimeter and allowed to cool in the same environment, then the rate of loss of heat of turpentine oil will be (Given, relative density of oil = 0.9).

Q. A calorimeter of water equivalent

5 \times 10^{- 3} k g

contains

25 \times 10^{- 3} k g

of water. It takes 3 minutes to cool from

28^{\circ} C

21^{\circ} C

. When the same calorimeter is filled with

30 \times 10^{- 3} k g

of turpentine oil, then it takes 2 minutes to cool from

28^{\circ} C

21^{\circ} C

. Find the specific heat of turpentine oil.

A calorimeter of negligible heat capacity contains 100 cc of water at $40^{0}$ C. The water cools to $35^{0}$ C in 5 minutes. The water is now replaced by K-oil of equal volume at $40^{0}$ C. Find the time taken for the temperature to become $35^{0}$ C under similar conditions. Specific heat capacities of water and K-oil are 4200 J $k g^{- 1} K^{- 1}$ and 2100 J $k g^{- 1} K^{- 1}$ respectively. Density of K-oil = 800 kg $m^{- 3}$ .

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