Question

A 52-gram sample of water that has an initial temperature of 10.0 °C absorbs 4,130 joules. If the specific heat of water is 4.184 J/(g °C), what is the final temperature of the water?

Answer 1

29 degree Celsius

Step-by-step explanation:

If anyone confused this is the choices:

11 °C

19 °C

29 °C

Estimate 28.98 to 29 for the answer

Answer 2

The final temperature of the water is 28.98 degree Celsius.

Step-by-step explanation:

It is given that,

Mass of sample of water, m = 52 grams

Initial temperature,
`T_i=10^(\circ)C`

Heat absorbed,
`Q=4,130\ J`

The specific heat of water is
`4.184\ J/(g^(\circ) C)`

We need to find the final temperature of the water. The heat absorbed is given by the formula as follows :

`Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\T_f=(Q)/(mc)+T_i\\\\T_f=(4130)/(52* 4.184 )+10\\\\T_f=28.98^(\circ) C`

So, the final temperature of the water is 28.98 degree Celsius.