F.3 PHYSICS

First, you should calculate the

1. Let c and P be the specific heat capacity of soup and the power of heater respectively.

Pt=mc^T
(15)(60)( P)=(1)(c)(70-25)

so, 20P=c --------(1)

2. Let t (seconds) be the time required.

Pt=(1.5)(c)(80-20)
Pt=(1.5)(20P)(80-20) (from (1))
t=1800s
=30mins

Heat supplied to heat 1 kg of soup from 25'C to 70'C
= 1 x C x (70-25) J = 45C J
where C is the specific heat capacity of the soup

Power delivered by the heater = 45C/15 J/min = 3C J/min

Heat needed to heat 1.5kg of soup from 20'C to 80'C
= 1.5C x (80-20) J = 90C J
Time required = 90C/3C minutes = 30 minutes