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  • A mass of 50g of water in a closed vessel, with surroundings at a constant temperature takes 2 minutes to cool from <img alt="30^{\circ}C\: \: to\: \: 25^{\circ}C." src="https://learn.careers3

A mass of 50g of water in a closed vessel, with surroundings at a constant temperature takes 2 minutes to cool from 30^{\circ}C\: \: to\: \: 25^{\circ}C. A mass of 100g of another liquid in an identical vessel with identical surroundings takes the same time to cool from 30^{\circ}C\: \: to\: \: 25^{\circ}C. The specific heat of the liquid is : 

(The water equivalent of the vessel is 30g.) 

Option: 1

2.0kcal/kg
 


Option: 2

7kcal/kg


Option: 3

3kcal/kg

 


Option: 4

0.5kcal/kg


Answers (1)

best_answer

 

 

 

 

As the surrounding is identical, the vessel is identical and the time taken to cool both water and liquid (from 300C to 250C) is same 2 minutes.

\begin{aligned} &\therefore\left(\frac{d Q}{d t}\right)_{\text {water }}=\left(\frac{d Q}{d t}\right)_{\text {liquid}}\\ &\text { Or } \frac{\left(m_{w} c_{w}+W\right) \Delta T_{1}}{t_{1}}=\frac{\left(m_{l} c_{l}+W\right) \Delta T_{2}}{t_{2}} \end{aligned}

where W=water equivalent of the vessel

and since \Delta T_{1}=\Delta T_{2}, t_{1}=t_{2}

we get 

m_{w} c_{w}=m_{l} c_{l}

∴ specific heat of liquid,

C_{l}=\frac{m_{w} c_{w}}{m_{l}}=\frac{50 \times 1}{100}=0.5 \mathrm{kcal} / \mathrm{kg}

 

Posted by

vishal kumar

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