Question

Water heated in an open pan where the air pressure is 10^5 pa. The water remains a liquid and

Answer 1

Answer:

Explanation:

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Answer 2

The ratio of the work done by the water to the heat absorbed by the water is $4.94 * 10^{-3}$

Explanation:

Given:

Pressure P = $10^{5}$ Pa

To find: The ratio

of the work done by the water to the heat absorbed by the water.

Solution

$P = 10^{5}$ Pa

W = Work done

Q = Heat absorbed

P = Air Pressure

C =specific heat of water

One calorie / gram is C = 4.186 joule/gram

β = Compressibility of water

β= 207 × $10^{-6}$

p = Density of water

p = 1000 $\frac{kg}{m^{3} }$

Calculate the ratio of work done and heat absorbed

$\frac{W}{Q} =$ PΔV/CmΔT

= Pβ/cp

= $\frac{1.00*10^{5}(207*10^{-6} ) }{4186.1}$

= $4.94 * 10^{-3}$

Final answer:

The ratio of the work done by the water to the heat absorbed by the water is $4.94 * 10^{-3}$

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