Question

How many joules are absorbed when 64.3 grams of water is heated from 40.0 °C to 74.0 °C?

A. 7,330 joules
B. 9,150 joules
C. 10,100 joules
D. 12,800 joules

Answer 1

Answer : The correct option is, (B) 9150 joules

Explanation : Given,

Initial temperature =
`40.0^oC`

Final temperature =
`74.0^oC`

Mass of water = 64.3 g

Specific heat of water =
`4.184J/g^oC`

Formula used :

`Q=m* C_w* \Delta T`

or,

`Q=m* C_w* (T_2-T_1)`

where,

Q = heat absorbed = ?

m = mass of water

`C_w` = specific heat of water

`T_1` = initial temperature

`T_2` = final temperature

Now put all the given value in the above formula, we get:

`Q=64.3g* 4.184J/g^oC* (74-40)^oC`

`Q=9150J`

Therefore, the heat absorbed will be, 9150 J