Question

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

Answer 1

Answer:

3966.82 J

Explanation:

q=sm∆T

q=73×13×4.18

the specific heat for water is 4.18

Answer 2

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.

$q= 39668.2 \ J$

39,668.2 Joules of heat energy are absorbed.