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

Answer:

The final temperature of the water is 28.98 degree Celsius.

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 :

$Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\T_f=\dfrac{Q}{mc}+T_i\\\\T_f=\dfrac{4130}{52\times 4.184 }+10\\\\T_f=28.98^{\circ} C$

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