
A 2100 W continuous flow geyser (instant geyser) has water inlet temperature 10 \[{}^\circ C\] while the water flows out at the rate of 20 g/s. The outlet temperature of the water must be about
A. 20\[{}^\circ C\]
B. 30\[{}^\circ C\]
C. 35\[{}^\circ C\]
D. 40\[{}^\circ C\]
Answer
565.2k+ views
Hint: Inside the geyser when water comes inside, energy is used to increase the temperature of the water. The time needed to heat the water is dependent upon the power wattage rating of the geyser.
Complete step by step answer:
The formula to be used here is \[Q=mc\Delta T\]
Here Q is the energy needed
c is the specific heat of the water, for water c=4200 \[Jk{{g}^{-1}}{{K}^{-1}}\]
\[\Delta T\] is the increment in temperature
Also \[\dfrac{Q}{t}\]= power=P
Given P= 2100W
\[\dfrac{m}{t}\]=20g/s= 0.02kg/s
Now, \[\dfrac{Q}{t}=\dfrac{m}{t}c\Delta T\]
On substituting values, we get
\[2100=0.02\times 4200\times \Delta T\]
\[\Delta T=\dfrac{2100}{0.02\times 4200}=25\]
\[\Delta T={{T}_{f}}-T\]
\[25={{T}_{f}}-10\]
\[35={{T}_{f}}\]
Hence the outlet temperature is 35\[{}^\circ C\]
So, the correct answer is “Option C”.
Note:
In this question, we have come to the relationship between power and increment in temperature of the water. a specific heat of a body is defined as the amount of energy required to increase the temperature of the body of mass 1g by 1\[{}^\circ C\]. So had it been some other liquid in the geyser the amount of energy would have been different. We come across different questions in which we have to use specific heat. Also not only for liquids but there exists specific heat of gases too. There exists specific heat at constant volume and specific heat at constant pressure. Not only this, but specific heat also changes with change in temperature, but for a small change in temperature specific heat remains constant.
Complete step by step answer:
The formula to be used here is \[Q=mc\Delta T\]
Here Q is the energy needed
c is the specific heat of the water, for water c=4200 \[Jk{{g}^{-1}}{{K}^{-1}}\]
\[\Delta T\] is the increment in temperature
Also \[\dfrac{Q}{t}\]= power=P
Given P= 2100W
\[\dfrac{m}{t}\]=20g/s= 0.02kg/s
Now, \[\dfrac{Q}{t}=\dfrac{m}{t}c\Delta T\]
On substituting values, we get
\[2100=0.02\times 4200\times \Delta T\]
\[\Delta T=\dfrac{2100}{0.02\times 4200}=25\]
\[\Delta T={{T}_{f}}-T\]
\[25={{T}_{f}}-10\]
\[35={{T}_{f}}\]
Hence the outlet temperature is 35\[{}^\circ C\]
So, the correct answer is “Option C”.
Note:
In this question, we have come to the relationship between power and increment in temperature of the water. a specific heat of a body is defined as the amount of energy required to increase the temperature of the body of mass 1g by 1\[{}^\circ C\]. So had it been some other liquid in the geyser the amount of energy would have been different. We come across different questions in which we have to use specific heat. Also not only for liquids but there exists specific heat of gases too. There exists specific heat at constant volume and specific heat at constant pressure. Not only this, but specific heat also changes with change in temperature, but for a small change in temperature specific heat remains constant.
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