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 dа ta:

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.