An ice cube of mass 0.1 kg at 0°C is placed in an isolated container at 227°C . Specific heat of container is given by S=A+BT, where A=100 cal kg−1K−1 and B=2×10−2cal kg−1K−2 . If final temperature after stabilization is 27°C , then which of these are correct ? (Take, Lwater=80 cal g−1 and Cwater=103 cal kg−1K−1 )

Heat received by ice =mL+mCΔT=10700cal Heat given by container = −∫500300mC(A+BT)dT =−mC[AT+BT22]500300=21600mC Heat lost = Heat gained ⇒10700=21600mC mC=1070021600=0.495⇒0.5 kg

An ice cube of mass 0.1 kg at 0°C is placed in an isolated container at 227°C . Specific heat of container is given by S=A+BT, where A=100 cal kg−1K−1 and B=2×10−2cal kg−1K−2 . If final temperature after stabilization is 27°C , then which of these are correct ? (Take, Lwater=80 cal g−1 and Cwater=103 cal kg−1K−1 )

Heat received by ice =mL+mCΔT=10700cal Heat given by container = −∫500300mC(A+BT)dT =−mC[AT+BT22]500300=21600mC Heat lost = Heat gained ⇒10700=21600mC mC=1070021600=0.495⇒0.5 kg

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An ice cube of mass 0.1 kg at $0 ° C$ is placed in an isolated container at $227 ° C$ . Specific heat of container is given by $S = A + B T,$ where $A = 100 cal k g^{- 1} K^{- 1}$ and $B = 2 \times 10^{- 2} cal k g^{- 1} K^{- 2}$ . If final temperature after stabilization is $27 ° C$ , then which of these are correct ? (Take, $L_{w a t e r} = 80 {cal g}^{- 1}$ and $C_{w a t e r} = 10^{3} c a l k g^{- 1} K^{- 1}$ )

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Ice receives a total of 10700 calories of heat

Container lost around 10700 calories of heat

Container has a mass of around 0.5 kg

Water equivalent of container is 300 g

answer is A, B, C.

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Detailed Solution

$H e a t r e c e i v e d b y i c e = m_{L} + m_{C} Δ T = 10700 c a l H e a t g i v e n b y c o n t a i n e r = - \int_{500}^{300} m_{C} (A + B T) d T = - m_{C} {[A T + \frac{B T^{2}}{2}]}_{500}^{300} = 21600 m_{C} H e a t l o s t = H e a t g a i n e d \Rightarrow 10700 = 21600 m_{C} m_{C} = \frac{10700}{21600} = 0.495 \Rightarrow 0.5 k g$