Question

Duncan knows that it takes 36400cal 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 0.750pint 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

So if he uses 75% of the original amount, then he uses: 36400 cal * 0.75 = 27300 cal

1 cal = 4.184 J, so 27300 cal*\(\dfrac{4.184~J}{1~cal}=114223.2~J\)

1 kJ = 1000 J, so \(114223.2~J*\dfrac{1~kJ}{1000~J}=114.2232~kJ\approx 114~ kJ \)