Question

How much energy must be absorbed by 45.0g of water to increase its temperature from 83 0°C to 303.0°C?

Answer 1

Answer: There will be 41382 J energy must be absorbed by 45.0g of water to increase its temperature from
`83.0^(o)C` to
`303.0^(o)C`.

Step-by-step explanation:

Given: Mass = 45.0 g

Initial temperature =
`83.0^(o)C`

Final temperature =
`303.0^(o)C`

Formula used to calculate heat energy is as follows.

`q = m * C * (T_(2) - T_(1))`

where,

q = heat energy

m = mass of substance

C = specific heat =
`4.18 J/g^(o)C` (here, for water)

`T_(1)` = initial temperature

`T_(2)` = final temperature

Substitute the values into above formula as follows.

`q = m * C * (T_(2) - T_(1))\\= 45.0 g * 4.18 J/g^(o)C * (303.0 - 83.0)^(o)C\\= 41382 J`

Thus, we can conclude that there will be 41382 J energy must be absorbed by 45.0g of water to increase its temperature from
`83.0^(o)C` to
`303.0^(o)C`.