Respuesta :
Answer : The correct option is, (c) [tex]3.7\times 10^2J/^oC[/tex]
Explanation :
First we have to calculate the energy or heat.
Formula used :
[tex]E=V\times I\times t[/tex]
where,
E = energy (in joules)
V = voltage (in volt)
I = current (in ampere)
t = time (in seconds)
Now put all the given values in the above formula, we get:
[tex]E=(3.6V)\times (2.6A)\times (350s)[/tex]
[tex]E=3276J[/tex]
Now we have to calculate the heat capacity of the calorimeter.
Formula used :
[tex]C=\frac{E}{\Delta T}=\frac{E}{T_{final}-T_{initial}}[/tex]
where,
C = heat capacity of the calorimeter
[tex]T_{initial}[/tex] = initial temperature = [tex]20.3^oC[/tex]
[tex]T_{final}[/tex] = final temperature = [tex]29.1^oC[/tex]
Now put all the given values in this formula, we get:
[tex]C=\frac{3276J}{(29.1-20.3)^oC}[/tex]
[tex]C=372.27J/^oC=3.7\times 10^2J/^oC[/tex]
Therefore, the heat capacity of the calorimeter is, [tex]3.7\times 10^2J/^oC[/tex]
The heat capacity of the calorimeter to calibrate it electrically, is the amount heat required to change its temperature to one degree. The heat capacity of calorimeter is = [tex]3.7\times10^2 \rm J/C^o[/tex].
What is heat capacity?
The heat capacity of a object is the property of it, due to which the object required a certain amount of heat to change its temperature of one degree.
It can be given as,
[tex]C=\dfrac{E}{t_f-t_i}[/tex]
Here, (E) is the energy produced and (t) is the temperature (subscript f used for final and i used for initial).
As the energy produced is the product of voltage (V) and current (I) and the time (t) of which it is given to the object. Thus the above formula can be written as,
[tex]C=\dfrac{VIt}{t_f-t_i}[/tex]
Given information-
The constant voltage required to calibrate the calorimeter is 3.6 V.
The current required to calibrate the calorimeter is 2.4 A.
The time of current flow is 350 seconds.
The initial temperature of the process is 20.3°C and final temperature is 29.1°C.
Put the value in the above formula as,
[tex]C=\dfrac{3.6\times2.4\times350}{29.1-20.3}\\C=3.7\times10^2\rm J/C^o[/tex]
Hence, the heat capacity of the calorimeter is [tex]3.7\times10^2 \rm J/C^o[/tex]
Learn more about the heat capacity here;
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