Question

The specific heat capacity of liquid water is 4.18 J/g oC. Calculate the quantity of energy required to heat 1.50 g of water from 26.5oC to 83.7oC. (Ignore significant figures for this problem.)

Answer 1

Answer: The quantity of heat required is 358.644 J.

Step-by-step explanation:

Given: Specific heat capacity =
`4.18 J/g^(o)C`

Mass = 1.50 g

`T_(1) = 26.5^(o)C`

`T_(2) = 83.7^(o)C`

Formula used to calculate heat energy is as follows.

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

where,

q = heat energy

m = mass

C = specific heat capacity

`T_(1)` = initial temperature

`T_(2)` = final temperature

Substitute the values into above formula as follows.

`q = m * C * (T_(2) - T_(1))\\= 1.50 * 4.18 J/g^(o)C * (83.7 - 26.5)^(o)C\\= 358.644 J`

Thus, we can conclude that quantity of heat required is 358.644 J.