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Duncan knows that it takes 36400 cal 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 0.850 pint 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 :

Answer:

129.453 kJ is required to heat 0.850 pints of water

Explanation:

1 pint requires 36400 cal of energy to heat a pint of water from room temperature to boil.

Sometimes he uses 0.850 pints. The energy he needs then : 36400 * 0.85 = 30940 cal

Since we know that 1 cal = 4.184 J

⇒ 1 cal = 4.184J  so 30940 cal =4.184 * 30940 = 129452.96 J = 129.453 kJ

So 0.850 pints of water require 129.453 kJ, at roomtemperature.

129.453 kJ is required to heat 0.850 pints of water.

Given:

1 pint requires 36400 cal of energy to heat a pint of water from room temperature to boil.

How to solve a problem?

Sometimes he uses 0.850 pints. The energy he needs then :

[tex]36400 * 0.85 = 30940 cal[/tex]

Since we know that [tex]1 cal = 4.184 J[/tex]

⇒  [tex]1 cal = 4.184 J[/tex]

So, [tex]30940 cal =4.184 * 30940 = 129452.96 J = 129.453 kJ[/tex]

So 0.850 pints of water require 129.453 kJ, at room temperature.

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