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Flura [38]
3 years ago
11

A 52-gram sample of water that has an initial temperature of 10.0 °C absorbs 4,130 joules. If the specific heat of water is 4.18

4 J/(g °C), what is the final temperature of the water?
Chemistry
1 answer:
FinnZ [79.3K]3 years ago
3 0

Answer:

The final temperature of the water is 28.98 degree Celsius.

Explanation:

It is given that,

Mass of sample of water, m = 52 grams

Initial temperature, T_i=10^{\circ}C

Heat absorbed, Q=4,130\ J

The specific heat of water is 4.184\ J/(g^{\circ} C)

We need to find the final temperature of the water. The heat absorbed is given by the formula as follows :

Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\T_f=\dfrac{Q}{mc}+T_i\\\\T_f=\dfrac{4130}{52\times 4.184 }+10\\\\T_f=28.98^{\circ} C

So, the final temperature of the water is 28.98 degree Celsius.

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The solubility of nitrogen gas in water is 1.90 mL/dL at 1.00 atm and 13.3 mL/dL at 7.00 atm.

We want to relate the solubility of a gas with its partial pressure.

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<h3>What does Henry's law state?</h3>

Henry's law states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid.

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