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 $303.0^{o}C$.

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.

$q = m \times C \times (T_{2} - T_{1})$

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.

$q = m \times C \times (T_{2} - T_{1})\\= 45.0 g \times 4.18 J/g^{o}C \times (303.0 - 83.0)^{o}C\\= 41382 J$

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 $303.0^{o}C$.