Question

If a hot 500mL bowl of soup was 88 degrees, then cooled down to 10 degrees, how much heat was lost?

Hint: not 88-10​
Can you help please!

Answer 1

163.02 kJ

Step-by-step explanation:

Heat energy can be calculated using

Q=mcθ

where Q = Heat energy

m =mass

c = specific heat capacity

θ =temperature difference

here we need the specific heat capacity and the density of the soup.

Assumptions

• specific heat be approximate to specific heat of water =4.18 J/g C
• density = 1 g/cm3 (water)
• 1 ml = 1 cm3
• mass= volume * density

Q=mcθ

Q=(v*d)*cθ

Q = (500* 1) * 4.18 *(10-88)

Q = -163020 J

Q= 163.02 kJ