An ice cube of mass 0-1 Kg at 0oC is placed in an isolated container which is at 227oC- The specific heat s of the container varies with temperature T according to the empirical relation s-A-BT- where A-100 cal kg-1K-1 and B-2-10-2 cal kg-1-If the final temperature of the container is 27o C- the mass of the container is Latent heat of fusion of water -8- 104 cal kg-1- Specific heat of water- 103 cal kg-1K-1

The correct option is C 0.495 kg Heat lost by container = ∫^300_500(A+BT)dT = m_c[AT+BT^22]^300_500 ⇒21600m_c Heat gained by the ice is: ...

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Byju's Answer

Standard XII

Physics

Specific Heat Capacity

An ice cube o...

Question

An ice cube of mass $0.1 K g$ at $0^{o} C$ is placed in an isolated container which is at $227^{o} C$ . The specific heat $s$ of the container varies with temperature $T$ according to the empirical relation $s = A + B T,$ where $A = 100 c a l k g^{- 1} K^{- 1}$ and $B = 2 \times 10^{- 2} c a l k g^{- 1}$ .
If the final temperature of the container is $27^{o} C$ , the mass of the container is (Latent heat of fusion of water $= 8 \times 10^{4} c a l k g - 1,$ Specific heat of water= $10^{3} c a l k g^{- 1} K^{- 1}$

0.495 k g

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0.595 k g

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0.695 k g

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0.795 k g

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Solution

The correct option is C $0.495 k g$
Heat lost by container $= - \int_{500}^{300} (A + B T) d T$

$= - m_{c} {[A T + \frac{B T^{2}}{2}]}_{500}^{300}$

$\Rightarrow 21600 m_{c}$

Heat gained by the ice is:
$= 0.1 \times 8 \times 10^{4} + 0.1 \times 10^{3} \times 27$
$= 10700 c a l$

According to principle of calorimetry
Heat lost by container = Heat gained by ice
$21600 m_{c} = 10700$
or $m_{c} = 0.495 k g$

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The correct option is C 0.495kgHeat lost by container =−∫300500(A+BT)dT=−mc[AT+BT22]300500⇒21600mcHeat gained by the ice is: =0.1×8×104+0.1×103×27 =10700calAccording to principle of calorimetryHeat lost by container = Heat gained by ice 21600mc=10700or mc=0.495kg