Heat Transfer

A container with

1%20%5C%2Ckg

of water in it is kept in sunlight, which causes the water to get warmer than the surroundings. The average energy per unit time per unit area received due to the sunlight is

700%5C%2C%20Wm%20%5E%7B-2%7D

and it is absorbed by the water over an effective area of

0.05%5C%2C%20m%20%5E%7B2%7D

.Assuming that the heat loss from the water to the surroundings is governed by Newton's law of cooling, the difference (in

%7B%20%7D%5E%7B%5Ccirc%7D%20C

) in the temperature of water and the surroundings after a long time will be _______ .(Ignore effect of the container, and take constant for Newton's law of cooling

%3D0.001%5C%2C%20s%20%5E%7B-1%7D

, Heat capacity of water

%3D4200%5C%2C%20J%20%5C%2Ckg%20%5E%7B-1%7D%20K%20%5E%7B-1%7D

)

container with 1 kg of water in it is kept

To solve the problem, we need to find the heat required to raise the temperature of 1 kg of a substance from $-73%5E%7B%5Ccirc%7DC$ to $27%5E%7B%5Ccirc%7DC$ .

1. Understanding the Specific Heat Capacity Relation:
The specific heat capacity of the substance is given by $C%20%3D%20kT$ , where $k$ is a constant and $T$ is the absolute temperature.

2. Converting Temperatures to Kelvin:
We have $T_1%20%3D%20-73%5E%7B%5Ccirc%7DC%20%3D%20-73%20%2B%20273%20%3D%20200%20%5C%2C%20%5Ctext%7BK%7D$ and $T_2%20%3D%2027%5E%7B%5Ccirc%7DC%20%3D%2027%20%2B%20273%20%3D%20300%20%5C%2C%20%5Ctext%7BK%7D$ .

3. Formula for Heat Required:
The heat required is given by:

$Q%20%3D%20%5Cint_%7BT_1%7D%5E%7BT_2%7D%20mC%20%5C%2C%20dT%20%3D%20%5Cint_%7BT_1%7D%5E%7BT_2%7D%20m(kT)%20%5C%2C%20dT$
where $m%20%3D%201$ kg.

4. Simplifying the Integral:
We now have:

$Q%20%3D%20%5Cint_%7B200%7D%5E%7B300%7D%201(kT)%20%5C%2C%20dT%20%3D%20k%20%5Cint_%7B200%7D%5E%7B300%7D%20T%20%5C%2C%20dT%20%3D%20k%20%5Cleft%5B%20%5Cfrac%7BT%5E2%7D%7B2%7D%20%5Cright%5D_%7B200%7D%5E%7B300%7D$

5. Calculating the Integral:
This becomes:

$Q%20%3D%20k%20%5Cleft(%5Cfrac%7B300%5E2%7D%7B2%7D%20-%20%5Cfrac%7B200%5E2%7D%7B2%7D%20%5Cright)%20%3D%20k%20%5Cleft(%5Cfrac%7B90000%7D%7B2%7D%20-%20%5Cfrac%7B40000%7D%7B2%7D%20%5Cright)$

6. Final Simplification:
We get:

$Q%20%3D%20k%20%5Cleft(%5Cfrac%7B50000%7D%7B2%7D%20%5Cright)%20%3D%20k%20(25000)$

7. Solving for $n$ :
Given that $Q%20%3D%20nk$ , we have:

$nk%20%3D%2025000k$
This implies:

$n%20%3D%2025000$

Final Answer:
The final answer is $%5Cboxed%7B25000%7D$ .

High-Yield Trend

Chapter Questions
2 MCQs

Official Solution

Official Solution

Heat Transfer

High-Yield Trend

Chapter Questions 2 MCQs

Official Solution

Official Solution

Chapter Questions
2 MCQs