Calculate how much energy (in joules) it took to heat the cool water up to the point at which it started to boil. remember, while we say waters boils at 100oc, that is only at sea level, so your water will probably boil at a lower temperature. don’t think your thermometer is wrong if your water is boiling at 97oc. it depends on your elevation and the weather on the day you are doing the experiment (high or low-pressure system in place). two cups of water have a mass of 473.18g, and the specific heat of water is 4.184 j/goc. show your work.

152443.454 J

Explanation:

The general equation to get the heat transfer is  

Q = m* cp* delta T  

Where:

Q = heat transfer (Joules, J)

m= mass of the substance (g)

Cp = specific heat (J/g°C)  

Delta T = change in temperature is (Tfinal -Tinitial)  

To be able to solve the equation we just need Delta T  

The final temperature is 97°C but we don’t know the initial temperature. They say that is cold water, not ice, so we can assume that the temperature over 0°C. Usually in the laboratory you work with standard conditions, which are standardized conditions for pressure and temperature that allows comparisons between data. For temperature the value is usually °C or 20°C this last one is used by the National Institute of Standards and Technology (NIST).

For this problem will use the standard condition for temperature. Since we know is cool water, we use 20°C for calculations.  

Delta T = 97°C – 20°C = 77°C

Now we can calculate the heat transfer.

Q = m* Cp* delta T  

m= 473.18g  

Cp = 4.184 J/g°C

Delta T = 77°C  

Q = 473.18 g * 4.184 J/g°C * 77°C = 152443.454 J or 152.443 kJ