Question

A 0.25-kg coffee mug is made from a material that has a specific heat capacity of 950 J/(kg · C°) and contains 0.30 kg of water. The cup and water are at 25° C. To make a cup of coffee, a small electric heater is immersed in the water and brings it to a boil in two minutes. Assume that the cup and water always have the same temperature and determine the minimum power rating of this heater.

Answer 1

The minimum power neccesary is:

`P=932,4 W`

Step-by-step explanation:

The water is bring to a boil so it goes from 25°C to 100°C, the temperature rise si:

`dT=100°C-25°C=75°C`

Considering that the cup is always at the same temperatura as the water the trasfered energy can be calculated as:

`Q=(m_(mug)*C_(p-mug)+m_(water)*C_(p-water))*dT`

We have that:

So:

`Q=(0,25kg*950 J/(kg.°C)+0,30kg*4181 J/(kg.°C))*75°C`

`Q=1491,8J/°C*75°C=111885 J`

Given that this energy was supplied in 2 minutes:

`t=2 min=120 sec`

The minimum power neccesary is:

`P=(Q)/(t)=932,4 W`