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The temperature of a liquid whose specific heat capacity varies with temperature as $C = 3+4T(J/kg.K)$ rises from $10^{o}C$ to $50^{o}C$ in $20$ minutes when a heater is used. If the heater generates $2430 \, J/min$ , calculate the mass (in Kg) of the water.
Option: 1
1

Option: 2
0.5

Option: 3
2

Option: 4
0

Answers (1)

best_answer

Heat energy is related to specific heat capacity as, $\mathrm{H=m\int_{T_1}^{T_2}\, CdT}$

Here $m$ is mass of the substance

$C =$ specific heat capacity of the substance

Given, $C = 3+4T$

$T_1 =10^{o}C = 283 K$

$T_2 =50^{o}C = 323 K$

Putting in equation,

$H=m\int_{283}^{323}\, (3+4T)dT$

$\Rightarrow H=m\left [ 3T+2T^{2} \right ]^{323}_{283}$

$\Rightarrow H=m\left [ 3(323-283)+2(323^{2}-283^{2}) \right ]$

$H=m\times 48600$

Heat generated by heater in $20$ minutes $= 2430\times20 = 48600 J$

Equating the heats,

$48600=m\times48600$

$m=1\, Kg$

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