An ice cube of mass 0-1 kg at 0-C is placed in an isolated container which is at 227-C- The specific heat S of the container varies with temperature T according to the empirical relation S - A - BT- where A - 100 cal- kg-K and B - 2- 10-2 cal-kg-K2- If the final temperature of the container is 27-C- determine the mass of the container- Latent heat of fusion of water - 8- 104 cal-kg-K- Specific heat of water - 103 cal-kg-K-

Heat received by ice is Q_1 = mL + mC T = 10700 cal Heat given by the container is Q_2 = ∫_500^300m_C (A + BT) dT = m_C [AT + BT^22 ]^300_500 = ...

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Standard XII

Physics

Calorimetry

An ice cube o...

Question

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

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Solution

Heat received by ice is
$Q_{1} = m L + m C △ T = 10700 c a l$
Heat given by the container is $Q_{2}$
$= - \int_{500}^{300} m_{C} (A + B T) d T = - m_{C} {[A T + \frac{B T^{2}}{2}]}_{500}^{300} = + 21600 m_{C}$
By principle of calorimetry, $Q_{1} = Q_{2}$
$\Rightarrow m_{C} = 0.495 k g$ .

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Heat received by ice isQ1=mL+mC△T=10700 calHeat given by the container is Q2=−∫300500mC(A+BT)dT=−mC[AT+BT22]300500=+21600 mCBy principle of calorimetry, Q1=Q2⇒mC=0.495 kg.

Heat received by ice is
$Q_{1} = m L + m C △ T = 10700 c a l$
Heat given by the container is $Q_{2}$
$= - \int_{500}^{300} m_{C} (A + B T) d T = - m_{C} {[A T + \frac{B T^{2}}{2}]}_{500}^{300} = + 21600 m_{C}$
By principle of calorimetry, $Q_{1} = Q_{2}$
$\Rightarrow m_{C} = 0.495 k g$ .