Question

A metal of mass 250g is heated to a temperature of 65°C. It is then placed in

50g of water at 20°C. Then final steady temperature of water becomes 25°C.

Neglecting the heat taken by the container, calculate the specific heat capacity

of the metal. Take specific heat capacity of water = 4.2 Jg-1K

-1

Answer 1

250*c*65-25=50*4.2*25-20
c=specific heat capacity of metal
250*40*c=50*4.2*5
10000c=250*4.2
C=250*4.2/10000
C=0.105 Jkg-1C-1

Answer 2

Answer:

The specific heat of metal is given by $0.105$ $Jg^{-1} K$ .

Explanation:

• When a body at a higher temperature is brought in contact with a body at a lower temperature than heat gained by the low-temperature body is equal to the heat lost by the higher temperature body.
• We will neglect the heat lost to the surroundings or taken by the container.

The formula for heat lost/gained is given by:

$H=M*s*(T2-T1)$

where, $M$ is the body mass.

            $s$ is the specific heat.

            $T1,T2$  are the initial and final temperatures.

Step 1:

Given metal mass $= 250 g$

          Temperature lost $= 65^{0} C-25^{0} C$

So, Heat lost by the metal $= 250 * s * (65^{0} C-25^{0} C)$

                                            $= 250 * s * 40$

Given water mass $= 50 g$

          specific heat $= 4.2 Jg^{-1} K$

         Temperature gained $= 25^{0} C-20^{0} C$

So, heat gained by the water $= 50 * 4.2 * (25^{0} C-20^{0} C)$

                                                 $= 1050 J$

Step 2:

As Heat lost $=$ Heat gained

$250 * s * 40=1050$

$10000 * s=1050$

$s=0.105$  $Jg^{-1} K$

So, the specific heat of metal is given by $0.105$ $Jg^{-1} K$ .