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The temperature of a liquid (specific heat capacity = $3000J/KgK$ ) rises from $10^{o}C$ to $50^{o}C$ in 2 minutes when a heater is used. If the heater generates $60000J/min$ , 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, $H=mC\Delta T$

Here $m$ is mass of the substance

$C=$ specific heat capacity of the substance

$\Delta T=$ change in temperature

Given, $\mathrm{C = 3000\, J/kg-K}$

$\mathrm{\Delta T=50^{o}C-10^{o}C=40^{o}C=40K}$

Putting in equation,

$\mathrm{H=m \times 3000 \times 40J}$

Heat generated by heater in 2 minutes

$\mathrm{=60000 \times 2=120000J}$

Equating the heats,

$120000=m \times120000$

$\mathrm{m=1\, kg}$

Posted by

Rishabh

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