3. How much energy is needed to raise 45 grams of water from 40°C to 115 °C?

Chemistry

1 answer:

Dafna1 [17]3 years ago

5 0

Answer:

Q = 114349.5 J

Explanation:

Hello there!

In this case, since this a problem in which we need to calculate the total heat of the described process, it turns out convenient to calculate it in three steps; the first one, associated to the heating of the liquid water from 40 °C to 100 °C, next the vaporization of liquid water to steam at constant 100 °C and finally the heating of steam from 100 °C to 115 °C. In such a way, we calculate each heat as shown below:

$Q_1=45g*4.18\frac{J}{g\°C}*(100\°C-40\°C)=11286J\\\\Q_2=45g* 2260 \frac{J}{g} =101700J\\\\Q_3=45*2.02\frac{J}{g\°C}*(115\°C-100\°C)=1363.5J$

Thus, the total energy turns out to be:

$Q_T=11286J+101700J+1363.5J\\\\Q_T=114349.5J$

Best regards!

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