Complete Question
A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.
(a) How much heat transfer is needed each second to raise the water temperature from 35.0°C to 100°C, boil it, and then raise the resulting steam from 100°C to 450°C? Specific heat of water is 4184 J/(kg · °C), the latent heat of vaporization of water is 2256 kJ/kg, and the specific heat of steam is 1520 J/(kg · °C).
J
(b) How much power is needed in megawatts? (Note: In real power plants, this process occurs under high pressure, which alters the boiling point. The results of this problem are only approximate.)
MW
Answer:
The heat transferred is [tex]Q = 5.866 * 10^9 J[/tex]
The power is [tex]P = 5866\ MW[/tex]
Explanation:
From the question we are told that
Mass of the water per second is [tex]m = 1917 \ kg[/tex]
The initial temperature of the water is [tex]T_i = 35^oC[/tex]
The boiling point of water is [tex]T_b = 100^oC[/tex]
The final temperature [tex]T_f = 450^oC[/tex]
The latent heat of vapourization of water is [tex]c__{L}} = 2256*10^3 J/kg[/tex]
The specific heat of water [tex]c_w = 4184 J/kg^oC[/tex]
The specific heat of stem is [tex]C_s =1520 \ J/kg ^oC[/tex]
Generally the heat needed each second is mathematically represented as
[tex]Q = m[c_w (T_i - T_b) + m* c__{L}} + m* c__{S}} (T_f - T_b)][/tex]
Then substituting the value
[tex]Q = m[c_w [T_i - T_b] + c__{L}} + C__{S}} [T_f - T_b]][/tex]
[tex]Q = 1917 [(4184) [100 - 35] + [2256 * 10^3] +[1520] [450 - 100]][/tex]
[tex]Q = 1917 * [3.05996 * 10^6][/tex]
[tex]Q = 5.866 * 10^9 J[/tex]
The power required is mathematically represented as
[tex]P = \frac{Q}{t}[/tex]
From the question [tex]t = 1\ s[/tex]
So
[tex]P = \frac{5.866 *10^9}{1}[/tex]
[tex]P = 5866*10^6 \ W[/tex]
[tex]P = 5866\ MW[/tex]
