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Nookie1986 [14]

2 years ago

12

What minimum heat is needed to bring 250 g of water at 20 ∘C to the boiling point and completely boil it away? The specific heat

of water is 4190 J/(kg⋅K) and its heat of vaporization is 22.6×105 J/kg.

1 answer:

Amiraneli [1.4K]2 years ago

4 0

Answer:633.8 KJ

Explanation:

Given

mass of water $\left ( m\right )=250gm$

Initial temperature $\left ( T_i\right )=20^{\circ}C$

Final temperature $\left ( T_f\right )=100^{\circ}C$

Specific heat of water $\left ( c \right )$ =4190 J/kg-k

heat of vaporization $\left ( L\right )=22.6\times 10^5 J/kg$

Heat required for process $\left ( Q\right )$ =heat to raise water temperature from 20 to 100 +Heat to vapourize water completely

Q=mc $\left ( T_f-T_i\right )+mL$

Q= $0.25\times 4190\times \left ( 100-20\right )+0.25\times 22\times 10^5$

Q= $\left ( 0.838+5.5\right )\times 10^5$

Q= $6.338\times 10^5J=633.8 KJ$

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