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4 years ago

7

An inventor develops a stationary cycling device by which an individual, while pedaling, can convert all of the energy expended

into heat for warming water.
How much mechanical energy is required to increase the temperature of 300 g of water (enough for 1 cup of coffee) from 20°C to 95°C?
(1 cal = 4.186 J, the specific heat of water is 4 186 J/kg⋅°C)
a. 94 000 J
b. 22 000 J
c. 5 400 J
d. 14 J

1 answer:

mr Goodwill [35]4 years ago

4 0

Answer:

$Q=94185\ J$

Explanation:

Given:

mass of water, $m=0.3\ kg$

initial temperature of water, $T_i=20^{\circ}C$

final temperature of water, $T_f=95^{\circ}C$

specific heat of water, $c=4186\ J.kg^{-1}.K^{-1}$

<u>Now the amount of heat energy required:</u>

$Q=m.c.\Delta T$

$Q=0.3\times 4186\times (95-20)$

$Q=94185\ J$

Since all of the mechanical energy is being converted into heat, therefore the same amount of mechanical energy is required.

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