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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 θ0=10.0°C and final temperature of calorimeter and unknown liquid is θf=73.0°C. Mass of empty calorimeter is m1=1.0 kg and combined mass of calorimeter and unknown liquid is m2=3.0 kg. Find the specific heat capacity of the unknown liquid in proper significant figures. Specific heat of calorimeter = 3.0×103 J/kg°C

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By Expert Faculty of Sri Chaitanya
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 1000 g12  cal g.C  -1 and 80°C respectively. The mass of the liquid and the calorimeter are 900 g and 200 g. Initially, both are at room temperature 20°C. Both calorimeter and the solid are made of same material. In the steady state, temperature of mixture is 40°C, then find the specific heat capacity of the unknown liquid.

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