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 $θ_{0} = 10.0 ° C$ and final temperature of calorimeter and unknown liquid is $θ_{f} = 73.0 ° C$ . Mass of empty calorimeter is $m_{1} = 1.0 kg$ and combined mass of calorimeter and unknown liquid is $m_{2} = 3.0 kg$ . Find the specific heat capacity of the unknown liquid in proper significant figures. Specific heat of calorimeter = $3.0 \times 10^{3} J / kg ° C$

Moderate

Solution

Given, $V = 100.0 V, i = 10.0 A, t = 700.0 s, θ_{0} = 10.0 ° C, θ_{f} = 73.0 ° C$ ,
$m_{1} = 1.0 kg$ and $m_{2} = 3.0 kg$
Substituting the values in the expression,
$S_{l} = (\frac{1}{m_{2} - m_{1}}) [\frac{V_{i t}}{θ_{f} - θ_{0}} - m_{1} S_{c}]$
$S_{l} = \frac{1}{3.0 - 1.0} [\frac{(100.0) (10.0) (700.0)}{73.0 - 10.0} - (1.0) (3.0 \times 10^{3})]$
$= 4.1 \times 10^{3} J / kg ° C$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

SCAN ME