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

Answer:

The minimum power neccesary is:

$P=932,4 W$

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: $m_{mug}=0,25kg, C_{p-mug}=950 J/(kg.°C), m_{water}=0,30kg, C_{p-water}=4180 J/(kg.°C) (considering it constant)$

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:

$P=\frac{Q}{t}=932,4 W$