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2 years ago

How many joules of heat are absorbed when 73 g water are heated from 30C to 43C? *

Chemistry

2 answers:

stepladder [879]2 years ago

6 0

Answer:

3966.82 J

Explanation:

q=sm∆T

q=73×13×4.18

the specific heat for water is 4.18

zysi [14]2 years ago

6 0

Answer:

$\boxed {\boxed {\sf 39,668.2 \ Joules}}$

Explanation:

We are given the mass and change in temperature, so we must use this formula for heat energy:

$q=mc \Delta T$

The mass is 73 grams. Water's specific heat is 4.18 J/g × °C. Let's calculate the change in temperature

ΔT= final temperature - initial temperature
ΔT= 43 °C - 30°C
ΔT= 13 °C

Now we know all the variables and can substitute them into the formula.

$m= 73 \ g \\c= 4.18 \ J/g* \textdegree C \\\Delta T= 13 \ \textdegree C$

$q= (73 \ g )(4.18 \ J/g*\textdegree C)(13 \textdegree C)$

Multiply the first numbers together. The grams will cancel.

$q= 3051.4 \ J/\textdegree C (13 \textdegree C)$

Multiply again. This time, the degrees Celsius cancel.

$q= 39668.2 \ J$

<u>39,668.2 Joules</u> of heat energy are absorbed.

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3 years ago

The solubility of nitrogen gas at 25 ◦C and 1 atm is 6.8×10−4 mol/L. If the partial pressure of nitrogen gas in air is 0.76 atm,

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Answer:

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3 years ago

Read 2 more answers