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2 years ago

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A 59.7 g piece of metal that had been submerged in boiling water was quickly transferred into 60.0 mL of water initially at 22.0

°C. The final temperature is 28.5°C. Use these data to determine the specific heat of the metal.

1 answer:

Anna007 [38]2 years ago

5 0

Answer:

$s = 382.45 J/kg C$

Explanation:

Here by energy equivalence we can say that energy given by the metal piece is same as the energy absorbed by the water

so here we have

$m_1s_2\Delta T_1 = m_2s_2 \Delta T_2$

here we know that

$m_1 = 60 mL = 0.060 kg$

$s_1 = 4186 J/kg C$

$\Delta T_1 = 28.5 - 22$

$m_2 = 59.7 g = 0.0597 kg$

$\Delta T_2 = 100 - 28.5$

$0.060(4186)(28.5 - 22) = (0.0597)(s)(100 - 28.5)$

$1632.54 = 4.27 s$

$s = 382.45 J/kg C$

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Answer:

A) 12.08 m/s

B) 19.39 m/s

Explanation:

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mg(sinθ) – F_k = ma

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Thus;

F_k = μ_k(mg cosθ)

We now have;

mg(sinθ) – μ_k(mg cosθ) = ma

Dividing through by m to get;

g(sinθ) – μ_k(g cosθ) = a

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a = -3.71 m/s²

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Using newton's 3rd equation of motion, we have;

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v = √(18² + (2 × -3.71 × 24))

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B) Now, μ_k = 0.10

Thus;

a = 9.8(sin 12.03) - 0.1(9.8 × cos 12.03)

a = 1.08 m/s²

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With your hand parallel to the floor and your palm upright, you lower a 3-kg book downward. If the force exerted on the book by

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For Newton's second law, the resultant of the forces acting on the book is equal to the product between the mass of the book and its acceleration:
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Two long straight wires enter a room through a window. One carries a current of 2.9A into the room, while the other carries a cu

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Answer and Explanation:

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i ' = 4.4 A

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Since from Ampere's law

where μ o = permeability of free space = 4π * 10 ^-7 H / m

plug the values we get the magnitude (in T.m) of the path integral of B.dl = ( 4π*10^-7 ) (2.9+4.4)

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Heat engines were first envisioned and built during the industrial revolution. Explain the thermodynamics of a heat engine comme

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Heat engines were developed during industrial revolution.

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Let us consider $T_{1}\ and\ T_{2}$ are the temperature of source and sink.

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$Let\ Q_{1}\ and\ Q_{2}$ are the heat provided by source and heat rejected to sink.

Hence, the work done by the working substance will be -

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Here output = workdone [W]

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This is the expression for the efficiency of heat engine.

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Answer:

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<u>Now, therefore the volume of ice:</u>

$V_i=V-V_o$

$V_i=90-60$

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<u>Now the volume of water:</u>

$V_w=V'-V_o$

$V_w=87-60$

$V_w=27\ cm^3$

As we know that the relative density is the ratio of density of the substance to the density of water.

<u>So, the relative density of ice:</u>

$S_i=\frac{\rho_i}{\rho_w}$ .....................(1)

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$\rm \rho=\frac{mass}{volume}$

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$S_i=\frac{m}{V_{i}}\div \frac{m}{V_w}$

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