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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!!!​

Respuesta :

Answer:

4000 [tex]\textbf{J}\text{ }\textbf{kg}^{\textbf{-1}}\text{ }\textbf{K}^{\textbf{-1}}[/tex]

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 [tex]s[/tex].

      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 = [tex]13600\text{ }J-(300\text{ }sec)\times(12\text{ }\frac{J}{sec})=10000\text{ }J[/tex]

       [tex]10000\text{ }J=m\times s\times\Delta T=(0.1\text{ }kg)\times(s)\times(323K-298K)=2.5s\text{ }kgK\\s=4000\text{ }\frac{J}{kg.K}[/tex]

∴ Specific heat capacity of liquid = 4000 [tex]\frac{J}{kg.K}[/tex]

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