How much heat (in kj) is needed to heat a 200.0 ml sample of water initially at 25.0 °c to a boiling temperature of 100.0 °c?

The amount of heat needed is 62.79 Kj.

Explanation:

Given data:

Mass of water = 200 mL

Initial Temperature = 25 °C

Final temperature = 100 °C

Amount of heat required = ?

Solution:

specific heat of water is 4.186 j/g.°C

Mass of water

1 mL = 1g

200 mL = 200 g

Change in temperature

ΔT = T2 - T1

ΔT = 100 - 25 = 75 °C

Formula:

Q = m. c. ΔT

Q = amount of heat

m = mass

c = specific heat

ΔT = change in temperature

Now we will put the values in formula,

Q = m. c. ΔT

Q = 200 g. 4.186 j/g°C . 75 °C

Q = 62790 j    

  62790/1000

Q = 62.79 Kj

The amount of heat needed is 62.79 Kj.

water and alcohols have similar properties because water molecules contain hydroxyl groups that can form hydrogen bonds with other water molecules and with alcohol molecules, and likewise alcohol molecules can form hydrogen bonds with other alcohol molecules as well as with water.

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