Answer:
The minimum power neccesary is:
[tex]P=932,4 W[/tex]
Explanation:
The water is bring to a boil so it goes from 25°C to 100°C, the temperature rise si:
[tex]dT=100°C-25°C=75°C[/tex]
Considering that the cup is always at the same temperatura as the water the trasfered energy can be calculated as:
[tex]Q=(m_{mug}*C_{p-mug}+m_{water}*C_{p-water})*dT[/tex]
We have that: [tex]m_{mug}=0,25kg, C_{p-mug}=950 J/(kg.°C), m_{water}=0,30kg, C_{p-water}=4180 J/(kg.°C) (considering it constant)[/tex]
So:
[tex]Q=(0,25kg*950 J/(kg.°C)+0,30kg*4181 J/(kg.°C))*75°C[/tex]
[tex]Q=1491,8J/°C*75°C=111885 J[/tex]
Given that this energy was supplied in 2 minutes:
[tex]t=2 min=120 sec[/tex]
The minimum power neccesary is:
[tex]P=\frac{Q}{t}=932,4 W[/tex]
