JEE MainPhysicsToughHeat

Temperature of 100 g of water in a thermoflaskremains fixed for
a pretty long time at 50ºC. Anequal mass of sand at 20ºC is poured in
the flaskand shaken for some time so that the temperatureof the mixture
is 40ºC. Now the experiment is repeated with 100 g of a liquid at 50ºC
and an equalamount of sand at 20ºC when the temperature ofthe mixture is
found to be 30ºC. The specific heat of the liquid (in
$kJ{kg}^{- 1}K^{- 1}$ ) is

Show answer and solution

m_{\omega}\ S_{\omega}\ \Delta T_{\omega} = m_{S}\ S_{S}\ \Delta T_{S}

100 \times 1 \times 10 = 100 \times S_{S} \times 20

\Rightarrow S_{S} = \frac{1}{2}cal/{gm}^{\circ}C

m_{S}\ S_{S}\ \Delta T_{S} = m_{\mathcal{l}}{\ S}_{\mathcal{l}}\ \Delta T_{\mathcal{l}}

100 \times \frac{1}{2} \times 10 = 100 \times {\ S}_{\mathcal{l}} \times 20

\Rightarrow S_{\mathcal{l}} = \frac{1}{4}cal/ºC\ gm

S_{\mathcal{l}} = \frac{4.2kJ}{4}/^{\circ}cgm = 1.05kJ/ºC\ gm

Source: Grass-G / P-08 Physics paper 23 October.docx

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