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.

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.

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

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.

$q = m \times C \times (T_{2} - T_{1})\\= 1.50 \times 4.18 J/g^{o}C \times (83.7 - 26.5)^{o}C\\= 358.644 J$

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