Question

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?

Answer 1

8856joules

Step-by-step explanation:

Energy= MC. ©

energy = 123.00*1*(97-25)

energy =8856joules//

Answer 2

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

Step-by-step explanation:

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

`Q=mc\Delta T`

where,

Q = heat absorbed or released

m = Mass of the substance = 123 kg

c = specific heat capacity of water =
`1J/kg.K`

`\Delta T` = change in temperature =
`(97-25)^oC=72^oC=72K` (change remains the same)

Putting values in above equation, we get:

`Q=123kg* 1J/kg.K* 72K\\\\Q=8856J`

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