When a block of metal of specific heat 0-1 cal-g- Cg and weighing 110 g is heated to 100- C and then quickly transferred to a calorimeter containing 200 g of a liquid at 10- C- the resulting temperature is 18- C- On repeating the experiment with 400 g of same liquid in the same calorimeter at same initial temperature- the resulting temperature is 14-5- C- If the specific heat of the liquid is x100 cal-g- C- Find - x-

Let s be the specific heat of the liquid and W be the water #160;equivalent of the calorimeter. Heat lost by the block = heat gained by ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Avogadro's Law & Gay Lussac's Law

When a block ...

Question

When a block of metal of specific heat $0.1 c a l / g^{\circ} C$ g and weighing $110 g$ is heated to $100^{\circ} C$ and then quickly transferred to a calorimeter containing $200 g$ of a liquid at $10^{\circ} C$ , the resulting temperature is $18^{\circ} C$ . On repeating the experiment with $400 g$ of same liquid in the same calorimeter at same initial temperature, the resulting temperature is ${14.5}^{\circ} C$ . If the specific heat of the liquid is $\frac{x}{100} c a l / g^{\circ} C$ . Find ' $x$ '.

Open in App

Solution

Let $s$ be the specific heat of the liquid and $W$ be the water
equivalent of the calorimeter.

$H e a t l o s t b y t h e b l o c k = h e a t g a i n e d b y (l i q u i d + c a l o r i m e t e r)$

For the first case:
$⟹ 110 \times 0.1 \times (100 - 18) = 200 \times s \times (18 - 10) + W \times 1 \times (18 - 10)$
$⟹ 1600 s + 8 W = 902$ (i)
For the second case:
$⟹ 110 \times 0.1 \times (100 - 14.5) = 400 \times s \times (14.5 - 10) + W \times 1 \times (14.5 - 10)$
$⟹ 1800 s + 4.5 W = 940.5$ (ii)
On solving equations (i) and (ii), we get $s = 0.48 c a l / g^{\circ} C$ and $W = 16.6 g$ .

Suggest Corrections

Similar questions

Q. Metal nails weighing

100 g

are heated to

100^{o} C

and dropped into a copper calorimeter of mass

100 g

containing

40 g

of water at

1^{o} C

. If the resulting temperature is

10^{o} C

, the specific heat of metal is (Sp. heat of copper is

0.1 c a l g^{- 1}^{o} C^{- 1}

Q. Two identical calorimeters 'X' and 'Y', of water equivalent 10 g each, contain equal quantities of the same liquid. The mass, initial temperature and specific heat capacity of the liquid present in both calorimeters are 50 g,

25^{\circ} C

and

\frac{1}{2} c a l g^{- 1}^{\circ} C^{- 1}

respectively. A 10 g metal piece (say A) of specific heat capacity

0.4 c a l g^{- 1}^{\circ} C^{- 1}

is dropped in calorimeter "X". and 25 g of metal piece (say B) of specific heat capacity "

S_{B}

" is dropped into calorimeter "Y". Due to this, the equilibrium temperature in 'X' and 'Y' rises to

30^{\circ} C

and

40^{\circ} C

respectively. If the initial temperature of 'B' is twice that of 'A', then find the value of "

S_{B}

Q. A piece of iron weighing

100

g was heated to

{98.5}^{o}

C and dropped in a calorimeter weighing

46

g and containing

85.4

g water at

15^{o}

C. The temperature was found to be

22^{o}

C. Calculate the specific heat of iron if the specific heat of material of calorimeter is

0.1

Q. A calorimeter contains

400 g

of water at a temperature of

5^{\circ} C

. Then,

200 g

of water at a temperature of

+ 10^{\circ} C

and

400 g

of ice at a temperature of

- 60^{\circ} C

are added. What is the final temperature(in

^{o} C

) of the contents of the calorimeter? Specific heat capacity of water

= 1000 c a l / k g / K

. Specific latent heat of fusion of ice

= 80 \times 1000 c a l / k g

. Relative specific heat of ice

= 0.5

Q. A metal object weighing

25^{\circ} C

is dropped in a calorimeter of mass

80

g and specific heat capacity of

1 J {k g}^{- 1} K^{- 1}

, containing 100 g of water at

40^{\circ} C

. final temperature recorded was

35^{\circ} C

. find the specific heat capacity of metal object.
specific heat capacity of water

4200 J {k g}^{- 1} K^{- 1}

Join BYJU'S Learning Program