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 -1 K -1 and B -2 - 10-2 cal kg -1- If the final temperature of the container is 27- C- then the mass of the container is Latent heat of fusion of water -8 - 104 calkg -1- specific heat of water -103 calkg -1 K -1 - A- 0-495 kgB- 0-595 kgC- 0-695 kgD- 0-795 kg

The correct option is A 0.495 kgHeat lost by container = ∫^300_500 m_c )(A+BT)dT = m_c [AT+BT^2/2 ]^300_500=21600m_c Heat gained by ice = m_ice L + m_ice S_wat ...

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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 $0^{\circ}$ 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 $A = 100 c a l k g^{- 1} K^{- 1}$ and $B = 2 \times 10^{- 2} c a l k g^{- 1}$ . If the final temperature of the container is $27^{\circ}$ C, then 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 $= 103 c a l k g^{- 1} K^{- 1})$ .

0.495 kg

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0.595 kg

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0.695 kg

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0.795 kg

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Solution

The correct option is A 0.495 kg
Heat lost by container = $- \int_{500}^{300} m_{c}) (A + B T) d T$
$= - m_{c} {[A T + \frac{B T^{2}}{2}]}_{500}^{300} = 21600 m_{c}$
Heat gained by ice $= m_{i c e} L + m_{i c e} S_{w a t e r} Δ T$
$= 0.1 \times 8 \times 104 + 0.1 \times 103 \times 27 = 10700 c a l$
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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Q. An ice cube of mass

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is placed in an isolated container which is at

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and

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The correct option is A 0.495 kgHeat lost by container =−∫300500mc)(A+BT)dT =−mc[AT+BT22]300500=21600mc Heat gained by ice =miceL+miceSwaterΔT =0.1×8×104+0.1×103×27=10700 cal According to principle of calorimetry Heat lost by container = Heat gained by ice 21600mC=10700 Or mC=0.495 kg.