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# In electrical calorimeter experiment, voltage across the heater is  and current is . Heater is switched on for . Room temperature is ${\theta }_{0}=10.0°\mathrm{C}$ and final temperature of calorimeter and unknown liquid is ${\theta }_{f}=73.0°\mathrm{C}$. Mass of empty calorimeter is  and combined mass of calorimeter and unknown liquid is . Find the specific heat capacity of the unknown liquid in proper significant figures. Specific heat of calorimeter =

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a
2.1×103 J/kg°C
b
4.1×103 J/kg°C
c
6.1×103 J/kg°C
d
8.1×103 J/kg°C
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detailed solution

Correct option is B

Given, V=100.0 V, i=10.0 A, t=700.0 s, θ0=10.0°C, θf=73.0°C,m1=1.0 kg and m2=3.0 kgSubstituting the values in the expression,Sl=1m2-m1Vitθf-θ0-m1ScSl=13.0-1.0(100.0)(10.0)(700.0)73.0-10.0-(1.0)3.0×103=4.1×103 J/kg°C

Similar Questions

The mass, specific heat capacity and the temperature of a solid are $\frac{1}{2}$   and $80°\mathrm{C}$ respectively. The mass of the liquid and the calorimeter are  and . Initially, both are at room temperature $20°\mathrm{C}$. Both calorimeter and the solid are made of same material. In the steady state, temperature of mixture is $40°\mathrm{C}$, then find the specific heat capacity of the unknown liquid.

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