Question

Compare the energy required to warm 1kg of water from 0ºC to 100ºC, to energy required to turn 1kg of water from liquid to vapor at 100ºC. a) warming the water requires more energy

b) vaporizing the water requires more energy
c) both processes require the same amount of energy

Answer 1

Answer: The correct answer is Option a.

Step-by-step explanation:

• The chemical equation for warming of water from 0°C to 100°C follows:

`H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)`

To calculate the amount of heat absorbed, we use the equation:

`q_1=m* C_(p,l)* (T_(2)-T_(1))`

where,

`q_1` = amount of heat absorbed = ?

`C_(p,l)` = specific heat capacity = 4.186 J/g °C

m = mass of water = 1 kg = 1000 g (Conversion factor: 1 kg = 1000 g)

`T_2` = final temperature =
`100^oC`

`T_1` = initial temperature =
`0^oC`

Putting all the values in above equation, we get:

`q_1=1000g* 4.186J/g^oC* (100-0)^oC=418600J`

• The chemical equation for vaporization of water follows:

`H_2O(l)\rightarrow H_2O(g)`

To calculate the amount of heat released, we use the equation:

`q_2=m* L_v`

where,

`q_2` = amount of heat absorbed = ?

m = mass of water = 1 kg

`L_f` = latent heat of vaporization = 2260000 J/kg

Putting all the values in above equation, we get:

`q_2=1kg* 2260000J/kg=2260000J`

From the calculations above, we get that
`q_1>q_2`

So, the warming of water requires more energy.

Hence, the correct answer is Option a.