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3 years ago

How much water

Physics

1 answer:

nevsk [136]3 years ago

6 0

Answer:

The mass is $m = 3.75 \ kg$

Explanation:

From the question we are told that

The initial temperature is $T_1 = 15^oC$

The final temperature is $T_2 = 100 ^o C \ (boiling point )$

Generally the maximum heat produced by 1 Liter of natural gas is $9000 \ cal$

So the amount of heat produced by 100 L is

$E = 9000 * 100$

=> $E = 9000 00 \ cal$

Generally given that the efficiency is $\eta = 0.35$

Then actual heat received by the water is

$H = 0.35 * E$

=> $H = 0.35 * 9000 00$

=> $H = 315000 \ cal$

Converting to kcal

=> $H = 315000 \ cal = \frac{315000}{1000} = 315 \ kcal$

Generally the specific heat of water is

$c_w = 1 kcal/ kg \cdot ^oC$

Generally the heat received by the water is mathematically represented as

$H = m * c_w * (T_2 - T_1)$

=> $315 = m * 1 * ( 100 - 15 )$

=> $m = 3.75 \ kg$

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