Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Experience the convenience of getting accurate answers to your questions from a dedicated community of professionals. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

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

The specific heat capacity of ice is 2.06 J/g °C

The specific heat capacity of water is 4.18 J/g °C

The specific heat capacity of steam is 2.02 J/g °C

The heat of fusion of water is 334 J/g

The heat of vaporization of water is 2260 J/g.


Sagot :

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:

[tex]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[/tex]

Thus, the total energy turns out to be:

[tex]Q_T=11286J+101700J+1363.5J\\\\Q_T=114349.5J[/tex]

Best regards!