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Umnica [9.8K]
3 years ago
9

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 ene

rgy 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!!!​
Physics
1 answer:
klemol [59]3 years ago
4 0

Answer:

<h2>4000 \textbf{J}\text{ }\textbf{kg}^{\textbf{-1}}\text{ }\textbf{K}^{\textbf{-1}} </h2>

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)\times(12\text{ }\frac{J}{sec})=10000\text{ }J

       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}

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

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The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing y_{1} is the same as the altitude y_{0} at which this projectile was launched: y_{0} = y_{1}.

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