Question

How much heat must be absorbed by 375 grams of water to raise its temperature by 25 degrees Celsius? Specific heat for water= 4.184 J/g/C or K

Answer 1

Answer:

$3.9*10^4 J$

Explanation:

• See image for explanation!

Answer 2

Answer:

$\boxed {\boxed {\sf 39, 225 \ Joules}}}$

Explanation:

Since we are given the mass, temperature, and specific heat, we should use the following formula to calculate heat energy.

$q=mc \Delta T$

We have 375 grams of water, the specific heat of water is 4.184 J/g ° C, and the temperature is raised 25 degrees Celsisus. Therefore:

• m= 375 g
• c= 4.184 J/g °C
• ΔT= 25°C

Substitute the values into the formula.

$q= (375 \ g)(4.184 \ J/g \textdegree C)(25 \textdegree C)$

Multiply the first two values together. The units of grams will cancel.

$q=1569 \ J/ \textdegree C(25 \textdegree C)$

Multiply again. This time, the degrees Celsius cancel, so we are left with only the units of Joules.

$q= 39225 \ J$

39,255 Joules of heat must be absorbed by the water.