Question

If a 375 mL sample of water was cooled from 37.5 degrees Celsius to 0 degrees Celsius , how much heat was lost (in joules)

Answer 1

Please, you have to apply the formula below:Q=c∗m∗Δtwhere Q is the energy lost, c is the specific heat of water, m is the mass of water involved, so m=3.75 *10^-1 Kg c=4,184 J/(Kg*°C) delta t=37.5 °C

Taking density of water as 1000kg/m3. Mass of water would be 0.375kg. So, heat lost would beH=mCDeltaTH=0.375*4184*37.5 = 58837.5J

Answer 2

Explanation:

It is given that volume is 375 mL and $\Delta T$ is 37.5 degree celsius.

First, we will calculate the mass of sample of water as it is known that density of water is 1 $g/ml$.

Hence,          Density = $\frac{mass}{volume}$

                   1 $g/ml$ = $\frac{mass}{375 ml}$

                           mass = 375 g

Therefore, formula to calculate heat loss will be as follows.

                         q = mC $\Delta T$

where          q = heat lost or absorbed

                    m = mass

                    C = specific heat of water = 4.186 J/g $^{o}C$

  $\Delta T$ = change in temperature

Hence, we will calculate value of heat lost as follows.

                      q = mC $\Delta T$

                         = 375 g × 4.186 J/g $^{o}C$ × ( 0 - 37.5) degree celsius

                          = - 58865.62 joules

Thus, we can conclude that heat lost is -58865.62 joules.