Question

To bring 1.0kg of water from 25Cto 99C takes how much heat input?

Answer 1

Use q=(m)(C)(delta T)

Answer 2

Heat,
`Q=3.10* 10^5\ J`

Step-by-step explanation:

Given that,

Mass of water, m = 1 kg

Initial temperature,
`T_i=25^(\circ)C`

Final temperature,
`T_f=99^(\circ)C`

The specific heat of water is,
`c=4200\ J/kg^(\circ)C`

The heat released due to the change in temperature is given by :

`Q=mc\Delta T`

`Q=mc(T_f-T_i)`

`Q=1* 4200* (99-25)`

`Q=3.10* 10^5\ J`

So, the heat input is
`Q=3.10* 10^5\ J`. Hence, this is the required solution.