Question

How much heat is absorbed by a 112.5 g sample of water when it is heated from 12.5 °C to 92.1 °C? (Specific heat capacity of water is 4.184 J/g °C)

Answer 1

Answer:

$\boxed {\boxed {\sf37,467.72 \ Joules }}$

Explanation:

We are asked to find how much heat a sample of water absorbed. Since we are given the mass, temperature, and specific heat, we will use the following formula.

$q=mc \Delta T$

The mass (m) of the sample is 112.5 grams. The specific heat capacity of water (c) is 4.184 Joules per gram degree Celsius. The difference in temperature (ΔT) is found by subtracting the initial temperature from the final temperature.

• ΔT= final temperature - initial temperature

The water was heated from 12.5 degrees Celsius to 92.1 degrees Celsius.

• ΔT= 92.1 °C - 12.5 °C= 79.6°C

Now we know three variables and can substitute them into the formula,

• m= 112.5 g
• c= 4.184 J/g °C
• ΔT= 79.6 °C

$q= (112.5 \ g )(4.184 \ J/g \textdegree C)(79.6 \textdegree C)$

Multiply the first 2 numbers. Note the units of grams cancel.

$q= (112.5 \ g *4.184 \ J/g \textdegree C)(79.6 \textdegree C)$

$q= (112.5 *4.184 \ J/ \textdegree C)(79.6 \textdegree C)$

$q= (470.7 \ J/ \textdegree C)(79.6 \textdegree C)$

Multiply again. This time the units of degrees Celsius cancel.

$q= (470.7 \ J/ \textdegree C *79.6 \textdegree C)$

$q= (470.7 \ J *79.6 )$

$q= 37467.72 \ J$

37, 467.72 Joules of heat are absorbed by the sample fo water.