Answer
Verified
88.8k+ views
Hint: In this question use the given information to identify the given values and also remember that \[\dfrac{{dT}}{{dt}} = -{5^ \circ }C\] which means every hour temperature will decrease by ${5^ \circ }C$, use this information to approach the solution.
Complete step-by-step solution:
According to the given information we know a room where the present temperature is ${40^ \circ }C$
So, for freezing process we have to lower the room temperature and the rate at which the room temperature is lowered every hour is ${5^ \circ }C$ i.e. \[\dfrac{{dT}}{{dt}} = -{5^ \circ }C\]
To find what will be the room temperature 10 hours after the process begins
Present temperature = ${40^ \circ }C$
Since the change in rate of temperature is \[\dfrac{{dT}}{{dt}} = -{5^ \circ }C\] which means in each hour temperature will drop by ${5^ \circ }C$
Therefore, after 10 hours the temperature will be drop by $-{5^ \circ }C \times 10$
Change in temperature after 10 hours = ${50^ \circ }C$
Therefore, temperature after 10 hours= ${40^ \circ }C - {50^ \circ }C$
So, the room temperature 10 hours after the process starts = $ - {10^ \circ }C$.
Note: The trick behind these types of questions is first to identify the initial temperature and the rate of change in temperature then as we knew that we require the change in temperature after 10 hours so it is a basic concept that if in an hour the change in temperature is ${5^ \circ }C$ so after 10 hours the change in temperature will be the multiplication of rate of change in temperature per hour and the time after finding the change in temperature after 10 hours we can subtract the initial temperature with the change in temperature to find the final temperature required.
Complete step-by-step solution:
According to the given information we know a room where the present temperature is ${40^ \circ }C$
So, for freezing process we have to lower the room temperature and the rate at which the room temperature is lowered every hour is ${5^ \circ }C$ i.e. \[\dfrac{{dT}}{{dt}} = -{5^ \circ }C\]
To find what will be the room temperature 10 hours after the process begins
Present temperature = ${40^ \circ }C$
Since the change in rate of temperature is \[\dfrac{{dT}}{{dt}} = -{5^ \circ }C\] which means in each hour temperature will drop by ${5^ \circ }C$
Therefore, after 10 hours the temperature will be drop by $-{5^ \circ }C \times 10$
Change in temperature after 10 hours = ${50^ \circ }C$
Therefore, temperature after 10 hours= ${40^ \circ }C - {50^ \circ }C$
So, the room temperature 10 hours after the process starts = $ - {10^ \circ }C$.
Note: The trick behind these types of questions is first to identify the initial temperature and the rate of change in temperature then as we knew that we require the change in temperature after 10 hours so it is a basic concept that if in an hour the change in temperature is ${5^ \circ }C$ so after 10 hours the change in temperature will be the multiplication of rate of change in temperature per hour and the time after finding the change in temperature after 10 hours we can subtract the initial temperature with the change in temperature to find the final temperature required.
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
Other Pages
Velocity of car at t 0 is u moves with a constant acceleration class 11 physics JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main
Formula for number of images formed by two plane mirrors class 12 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
A passenger in an aeroplane shall A Never see a rainbow class 12 physics JEE_Main