It’s a dynamic load.
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The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.
The given parameters;
- <em>initial temperature of metals, = </em>
<em /> - <em>initial temperature of water, = </em>
<em> </em> - <em>specific heat capacity of copper, </em>
<em> = 0.385 J/g.K</em> - <em>specific heat capacity of aluminum, </em>
= 0.9 J/g.K - <em>both metals have equal mass = m</em>
The quantity of heat transferred by each metal is calculated as follows;
Q = mcΔt
<em>For</em><em> copper metal</em><em>, the quantity of heat transferred is calculated as</em>;
<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>copper metal</em>;
<em>For </em><em>aluminum metal</em><em>, the quantity of heat transferred is calculated as</em>;
<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>aluminum metal </em><em>;</em>
Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.
Learn more here:brainly.com/question/15345295
Answer:
Total heat transfer is positive
Total work transfer is positive
Explanation:
The first law of thermodynamics states that when a system interacts with its surrounding, the amount of energy gained by the system must be equal to the amount of energy lost by the surrounding. In a closed system, exchange of energy with the surrounding can be done through heat and work transfer.
Heat transfer to a system is positive and that transferred from the system is negative.
Also, work done by a system is positive while the work done on the system is negative.
Therefore, from the question, since the heat engine inputs 10kJ of heat, then heat is being transferred to the system. Hence, the sign of the total heat transfer is positive (+ve)
Also, since the heat engine outputs 5kJ of work, it implies that work is being done by the system. Hence the sign of the total work transfer is also positive (+ve).
Answer: in this question, the only charge in the cavity is Q. Inside the conducting spherical shell, the electric field is zero.
While outside the shell, the electric field is given by: k(q + Q)/r²
Where;
K= is a constant which is given as, 8.99 x 10^9 N m² / C².
Q= source charge which creates the electric field
q= is the test charge which is used to measure the strength of the electric field at a given location.
r= is the radius
Explanation: Inside the conducting spherical shell, the electric field is zero since the Electric field vanishes everywhere inside the volume of a good conductor.
