Help me answer this!!

Question

A 106.3 mL sample of water has an initial temperature of 15.3°C. How many kilojoules (kJ) of heat must be absorbed to raise the temperature of the water to 33.0°C? The specific heat of water is 4.184 J/g·°C. Report your answer to 2 decimal places.

Answer 1

Hi Jenna,

To tackle this question, we use the concept of heat capacity or specific heat.

The amount of heat or energy (Q) needed to raise the temperature of any substance is as follows:

Q - m. C. ΔT where m is the mass of the substance, C is its specific heat and ΔT is the temperature rise.

In our case, we can simply plug in the values of what is given to us as follows:

Q = (106.3 g )(4.184 J/g.°C) (33 - 15.3) °C

Therefore, after multiplying, Q = 7872 J or 7.87 KJ

Note that I was able to substitute g for mL because water has a density of 1 g/mL.

Also, note that you always want to write the units in front of the numbers to avoid any errors in units.

Hope this helps.