Question

A mass of 50 g of a certain metal at 150°C is immersed in 100 g of water at 11°C. The final temperature is 20°C. Calculate the specific heat capacity of the metal. Assume that the specific heat capacity of water is :-
$\rm \: 4.2 \: Joule \: g^{ - 1} \: K ^{ - 1}$

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Answer 1

Answer:

A mass of 50 g of a certain metal at 150°C is immersed in 100 g of water at 11°C. The final temperature is 20°C.  Assume that the specific heat capacity of water is $\text { 4.2 Joule } \mathrm{g}^{-1} \mathrm{~K}^{-1}$ The specific heat capacity of the metal will be $0.58 \ \mathrm{Jg}^{-1} \mathrm{K}^{-1}$  

Explanation:

• We can solve this problem using the principle of calorie meter.

• Let M be the mass of water, m be the mass of the metal, T be the initial temperature of the water, t be the temperature of the metal, and θ be the final temperature.

       According to the principle of calorie meter,

       Heat gained by water =Heat lost by the metal

       That is,

       $\mathrm{MC}_{1}(\mathrm{~T}-\theta)=m \mathrm{C}_{2}(\theta-t)$

       Where

       $C_1$ - Specific heat capacity of water

       $C_2$ - Specific heat capacity of metal

In the question, it is given that,

$M = 50 g$          $m=100 \mathrm{~g}$    

$T = 150^0C$       $t = 11^0C$    

$\theta = 20^0C$         $C_1 = \text { 4.2 Joule } \mathrm{g}^{-1} \mathrm{~K}^{-1}$

Substitute these values into the above equation,

$100 \times 4.2(293-284)=50 \times \mathrm{C}_{\mathrm{s}} \times(423-293)$

$C_{2}=\frac{(420 \times 9) }{(50 \times 130)} =0.58 \ \mathrm{Jg}^{-1} \mathrm{K}^{-1}$

Answer 2

Answer:

• 0.5815 J/gK is the required specific heat capacity of metal

Explanation:

Topic

• Thermodynamics

According to the Question

It is given that ,

• Mass of metal ,m = 50g
• Initial temp of metal ,Ti = 150°C
• Final temperature ,Tf = 20°C
• Mass of water ,m' = 100g
• Initial temp of water ,Ti' = 11°C
• Specific heat capacity of water,C = 4.2

We have to calculate the specific heat capacity of metal .

Conversion of Unit

→ Initial temp of metal ,Ti = 150+273 = 423K

→ Final temp ,Tf = 20+ 273 = 293K

→ Initial temp of water ,Ti' = 11+273 = 284K

Calculating Change in Temperature :

For metal

→∆T = 423K - 293K = 130K

For Water

→ ∆T' = 293K - 284K = 9K

Now,

Let the specific heat capacity of metal be C .

For calculating specific heat capacity of metal we. will use here Principle of Calorie Meter which states that the heat energy lost by solid body is equal to heat gained by the body ( Isolated system) .

Heat Lost by metal = Heat gained by water

• mC∆T = m'C'∆T'

by putting the value we get

→ 50 × C × 130 = 100 × 4.2 × 9

→ 6500 × C = 420 × 9

→ C = 3780/6500

→ C = 0.5815 J/gK

• Hence, the specific heat capacity of metal is 0.5815 J/gK .