Question

Alculate the amount of heat transferred when 710 grams of water warms from an initial temperature of 4.0 ºC to a final temperature of 25.0 ºC. The specific heat capacity of liquid water is 4.184 J/g ºC. Please express you answer as a number in units of kJ, and be careful about the sign (is the answer a positive or negative number?) since that shows the direction of heat flow.

Answer 1

The Quantity of heat require is
`0.062kJ`

Explanation:

This problem bothers on the topic of heat capacity.

Given data

mass of water m=
`710 grams`

converting from grams to kg we have =
`(710)/(1000) = 0.71kg`

initial temperature T1= 4°C

final temperature T2= 25°C

heat capacity of water =
`4.184 J/g`

The expression for the quantity of heat required is given has

`Q= mc(T2-T1)`

substituting our given data we have

`Q= 0.71*4.184*(25-4)\\Q= 2.97*(21)\\Q= 62.38J\\`

In kilograms we have to divide by
`1000`

`Q= (62.38)/(1000) = 0.062kJ`