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DanielleElmas [232]
3 years ago
12

Suppose that we replace the aluminum with a mystery metal and repeat the experiment in the video. As in the video, the mass of t

he metal is the same as that of the water. Room temperature is about 20∘C before the start of the experiment. The water heats up to 40∘C, and the mystery metal heats up to 80∘C. Compared to that of water, the heat capacity of our mystery metal is
Physics
1 answer:
katrin2010 [14]3 years ago
7 0

Answer: b) one-third as great.

Explanation:

The options include:

a) three times greater.

b) one-third as great.

c) half as great.

d) two times greater.

e) the same.

Since the water heats up to 40°C and the room temperature is about 20°C before the start of the experiment, heat absorbed will be: (40°C-20°C).= 20°C

Since mystery metal heats up to 80∘C and the room temperature is about 20°C before the start of the experiment, heat absorbed will be: (80°C-20°C).= 60°C.

Therefore, based on the calculation, when compared to that of water, the heat capacity of our mystery metal is (20/60) = 1/3 one-third as great.

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A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a lar
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Answer:

W= 4.89 KJ

Explanation:

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temperature of hot water T₁ = 100⁰C

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Temperature of cold ice T₂= 0⁰C

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The latent heat of ice LH= 334 KJ

The heat rejected by the engine Q= m .LH

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I need help with this question how to solve it for Brass and Cooper
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Take into account that density and relative density are given by:

\begin{gathered} \text{density}=\text{ mass/volume} \\ \text{relative density = density/density of water} \end{gathered}

Take into account that the volume associated to each of the given sustances in the table is determined by the Level Difference (because it is the change in the volume of the water of the recipient in which the substance is immersed).

The density of water in kg/m^3 is 1000 kg/m^3.

Due to the density must be given in kg/m^3, it is necessary to express the volumes of the table in m^3 and mass in kg, then, consider the following conversion factor:

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Then, you obtain the following results:

Brass:

\begin{gathered} 53.2g\cdot\frac{1kg}{1000g}=0.0532kg \\ 6ml\cdot\frac{1m^3}{1000000ml}=0.000006m^3 \\ \text{density}=\frac{0.0532kg}{0.000006m^3}\approx8866.67\frac{kg}{m^3} \\ \text{relative density=}\frac{(\frac{8866.66kg}{m^3})}{(1000\frac{kg}{m^3})}\approx8.87 \end{gathered}

Cooper:

\begin{gathered} 57.4g=0.0574kg \\ 6ml=0.000006m^3 \\ \text{density}=\frac{0.0574kg}{0.000006m^3}\approx9566.67\frac{kg}{m^3} \\ \text{relative density=}\frac{\frac{9566.67kg}{m^3}}{1000kg}=9.57 \end{gathered}

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1 year ago
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