Question

Duncan knows that it takes 36400 calcal of energy to heat a pint of water from room temperature to boiling. However, Duncan has prepared ramen noodles so many times he does not need to measure the water carefully. If he happens to heat 0.900 pintpint of room-temperature water, how many kilojoules of heat energy will have been absorbed by the water at the moment it begins to boil?

Answer 1

Explanation:

It is given that 36400 cal per pint is the heat necessary to boil water. Hence, heat required to boil 0.9 pint of water is as follows.

            $0.9 pint \times \frac{36400 cal}{1 \text{pint}}$

             = 32760 cal

As 1 cal = 4.184 J. Therefore, the amount of energy absorbed by the 0.9 pint of water from room temperature to boiling point is as follows.

              $32760 \times 4.184$ J

               = 137067.84 J

or,           = 137.067 kJ      (as 1 kJ = 1000 J)

Therefore, we can conclude that heat energy required to be absorbed by the water at the moment it begins to boil is 137.067 kJ.