Question

How much energy, in kilojoules, must you add to 3.00 quarts of olive oil to warm it from 23.0 °C to 100.0 °C? The specific heat of olive oil is 0.1.79 J/g·°C and its density is 0.916 g/mL

Answer 1

3.58*10^-4kJ

Step-by-step explanation:

Given

T1=23.0 °C

T2=100.0 °C

c=1.79 J/g·°C

Volume of olive =3 quarts

Converting to liter

1 quart =0.946353 L

3 quarts=x

Cross multiplying

x=0.946353*3

x=2.83L

To mL= 0.0028mL

We are given the density as 0.916 g/mL

Mass= density *volume

Mass= 0.916*0.0028

Mass=0.0026g

We know that the quantity of heat is expressed as

Q=mcΔT

Q=0.0026*1.79(23-100)

Q=0.004654*(77)

Q=0.358J

In kJ we divide 0.36 by 1000

=0.000358358

=3.58*10^-4kJ