Question

In an experiment, it took 300s to increase the temperature of 0.1kg of the liquid from 25°C to 50°C. During this period, the energy supplied by the heater was 13600J, and the apparatus was losing heat at an average rate of 12J/s. Assuming the heat capacity of the container can be ignored, calculate the value of the specific heat capacity of the liquid. Pls I need urgent answers!!!​

Answer 1

4000
`\textbf{J}\text{ }\textbf{kg}^{\textbf{-1}}\text{ }\textbf{K}^{\textbf{-1}}`

Step-by-step explanation:

The temperature of 0.1 kg of liquid rises from 25°C to 50°C in 300 sec. Energy of 13,600 J was supplied during this time. Appartus was losing energy at the rate of 12 J/sec.

Let us assume the Specific heat capacity as
`s`.

As there is no state change from liquid to gas, only Specific heat capacity is involved. Also, work done is approximately zero because volume does not change much. So,

Energy gained = Energy required to rise the temperature

Energy gained by liquid =
`13600\text{ }J-(300\text{ }sec)*(12\text{ }(J)/(sec))=10000\text{ }J`

`10000\text{ }J=m* s*\Delta T=(0.1\text{ }kg)*(s)*(323K-298K)=2.5s\text{ }kgK\\s=4000\text{ }(J)/(kg.K)`

∴ Specific heat capacity of liquid = 4000
`(J)/(kg.K)`