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In electrical calorimeter experiment, vo...

In electrical calorimeter experiment, voltage across the heater is `100.0 V` and current is `10.0 A`. Heater is switched on for t `= 700.0 s`. Room temperature is `theta_0 = 10.0^(@)C` and final temperature of calorimeter and unknown liquid is `theta_f = 73.0^(@) C`. Mass of empty calorimeter is `m_1 = 1.0kg` and combined mass of calorimeter and unknown liquid is `m_2 = 3.0 kg`. Find the specificheat capacity of the unknown liquid in proper significant figures. Specific heat of calorimerter = `3.0xx10^3 j//kg .^(@) C`.

Text Solution

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The correct Answer is:
A, C, D
Given, `V= 100.0 V, i=10.0A, t=700.0 s,theta_0=10.0^(@) C,theta _f=73.0^(@) C,`
`m_1=1.0kg` and `m_(2)=3.0kg`
Substituting the values in the experssion.
`S_(l)=(1/(m_(2)-m_(1)))[(Vit)/(theta_(f)-theta_(0))-m_(1)S_(c)]`
we have, `S_(l)=1/(3.0-1.0)[((100.0)(10.0)(700.0))/(73.0-10.0)-(1.0)(3.0xx10^(3))]`
`= 4.1xx 10^(3) j//kg .^(@) C`
(According to the rules of significant figures).
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