Answer: 1896.55J/kg°C
Explanation:
The quantity of Heat Energy (Q) required to heat a material depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Since,
Q = 1320 joules
Mass of material = 5.61kg
C = ? (let unknown value be Z)
Φ = 0.124°C
Then, Q = MCΦ
1320J = 5.61kg x Z x 0.124°C
1320J = 0.696kg°C x Z
Z = (1320J / 0.696kg°C)
Z = 1896.55 J/kg°C
Thus, the specific heat of the material is 1896.55J/kg°C
The amount of diffraction depends on the wavelength of light, with shorter wavelengths being diffracted at a greater angle than longer ones (in effect, blue and violet<span> light are diffracted at a larger angle than is red light).
I hope my answer has come to your help. God bless and have a nice day ahead!
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Answer:
b. 9.5°C
Explanation:
= Mass of ice = 50 g
= Initial temperature of water and Aluminum = 30°C
= Latent heat of fusion = 
= Mass of water = 200 g
= Specific heat of water = 4186 J/kg⋅°C
= Mass of Aluminum = 80 g
= Specific heat of Aluminum = 900 J/kg⋅°C
The equation of the system's heat exchange is given by
The final equilibrium temperature is 9.50022°C
The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.
<h3>How does test charge affect electric field?</h3>
As the quantity of authority on the test charge (q) is increased, the force exerted on it is improved by the same factor. Thus, the ratio of force per charge (F / q) stays the same.
Adjusting the amount of charge on the test charge will not change the electric field force.
<h3>What is a test charge used for?</h3>
The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.
To learn more about test charge, refer
brainly.com/question/16737526
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