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Sergio [31]
2 years ago
9

Calculate the energy gained by an ice block in the following experiment.

Physics
1 answer:
MAXImum [283]2 years ago
7 0

The heat is exchanged when two different temperature objects come in contact. The energy gained by an ice block is 2.3 Joules.

<h3>What is temperature?</h3>

Temperature is the degree of hotness and coldness of the object.

A 7g block of ice was added to a coffee cup full of 103.4 grams of water. The water had an initial temperature T₁ = 24.5 C and a final temperature T₂ = 19.2 C after all the ice had melted.

Heat lost by water = Heat gained by ice

Qgain = ms(T₂ -T₁ )

Substituting the value for mass of water m =103.4 g= 0.1034 kg , specific heat of water s = 4.18 kJ/kg and temperature values, we get

Qgain = 0.1034 x 4.18 x (24.5 - 19.2)

Qgain = 2.3 Joules

Thus, the energy gained by an ice block is 2.3 Joules.

Learn more about temperature.

brainly.com/question/20459283

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Explanation:

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A thin, metallic spherical shell of radius 0.347 m0.347 m has a total charge of 7.53×10−6 C7.53×10−6 C placed on it. A point cha
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E = 12640.78 N/C

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In order to calculate the electric field you can use the Gaussian theorem.

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Q: net charge inside the Gaussian surface

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If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m

Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C

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E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}

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