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An sample of water with a mass of 123.00 kg is heated from 25 C to 97 C. If the specific heat of water is 1 J-1 kg K-1 then how much energy is required?

Respuesta :

Answer:

8856joules

Explanation:

Energy= MC. ©

energy = 123.00*1*(97-25)

energy =8856joules//

Answer: The amount of energy required will be 8856 Joules.

Explanation:

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

[tex]Q=mc\Delta T[/tex]

where,

Q = heat absorbed or released

m = Mass of the substance = 123 kg

c = specific heat capacity of water = [tex]1J/kg.K[/tex]

[tex]\Delta T[/tex] = change in temperature = [tex](97-25)^oC=72^oC=72K[/tex]  (change remains the same)

Putting values in above equation, we get:

[tex]Q=123kg\times 1J/kg.K\times 72K\\\\Q=8856J[/tex]

Hence, the amount of energy required will be 8856 Joules.