Respuesta :
Answer: The correct answer is Option a.
Explanation:
- The chemical equation for warming of water from 0°C to 100°C follows:
[tex]H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)[/tex]
To calculate the amount of heat absorbed, we use the equation:
[tex]q_1=m\times C_{p,l}\times (T_{2}-T_{1})[/tex]
where,
[tex]q_1[/tex] = amount of heat absorbed = ?
[tex]C_{p,l}[/tex] = specific heat capacity = 4.186 J/g °C
m = mass of water = 1 kg = 1000 g (Conversion factor: 1 kg = 1000 g)
[tex]T_2[/tex] = final temperature = [tex]100^oC[/tex]
[tex]T_1[/tex] = initial temperature = [tex]0^oC[/tex]
Putting all the values in above equation, we get:
[tex]q_1=1000g\times 4.186J/g^oC\times (100-0)^oC=418600J[/tex]
- The chemical equation for vaporization of water follows:
[tex]H_2O(l)\rightarrow H_2O(g)[/tex]
To calculate the amount of heat released, we use the equation:
[tex]q_2=m\times L_v[/tex]
where,
[tex]q_2[/tex] = amount of heat absorbed = ?
m = mass of water = 1 kg
[tex]L_f[/tex] = latent heat of vaporization = 2260000 J/kg
Putting all the values in above equation, we get:
[tex]q_2=1kg\times 2260000J/kg=2260000J[/tex]
From the calculations above, we get that [tex]q_1>q_2[/tex]
So, the warming of water requires more energy.
Hence, the correct answer is Option a.
