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

Step-by-step 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 =
`(mass)/(volume)`

1
`g/ml` =
`(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.