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nalin [4]
3 years ago
9

An engine is designed to obtain energy from the temperature gradient of the ocean. What is the thermodynamic efficiency of such

an engine if the temperature of the surface of the water is 59°F (15°C) and the temperature well below the surface is 41°F (5°C)
Physics
1 answer:
rewona [7]3 years ago
6 0

Answer:

0.035 (3.5 %)

Explanation:

The thermodynamic efficiency is given by:

\eta = 1 - \frac{T_C}{T_H}

where

T_C is the cold temperature

T_H is the hot temperature

In this problem we have

T_C = 5 ^{\circ}C+ 273 = 278 KT_H = 15^{\circ}C+273 = 288 K

So the efficiency is

\eta = 1 - \frac{278 K}{288 K}=0.035

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The heat (energy) needed to raise the temperature of the water is given by
Q=m C_S (T_f - T_i)=(350.0 g)(4.18 J/gC)(80C-20C)=87780 J

The wavelength of the radiation of the oven is \lambda=0.125 m, so the energy of a single photon of this radiation is
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3 years ago
A roadrunner is running along a straight desert road at a constant velocity of 25 m/s. If a certain coyote wants to capture the
andreyandreev [35.5K]

Answer:

t = 1.42 s and d = 35.5 m

Explanation:

Given that,

Velocity of a roadrunner is 25 m/s

A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.

We need to find the time before the roadrunner is under the overpass and  how far away from the overpass is the roadrunner when the coyote drops the net.

d=ut+\dfrac{1}{2}at^2\\\\\text{Here, u = 0 and a = g}\\\\d=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2d}{g}} \\\\t=\sqrt{\dfrac{2\times 10}{9.8}} \\\\t=1.42\ s

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d = vt

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