Question

A 6.60 g sample of solid KCl was dissolved in 47.6 g of water. The initial temperature of the water was 20.00°C. After the compound dissolved, the temperature of the water was 11.00°C. Assume the heat was completely absorbed from the water and no heat was absorbed by the reaction container or the surroundings. Calculate the heat absorbed by the process. The specific heat of water is 4.184 J/g·°C.

Answer 1

Answer : The heat absorbed by the process is 1792.4 J

Explanation :

In this problem we are given that heat was completely absorbed from the water and no heat was absorbed by the reaction container or the surroundings.

$q_{absorbed}=-q_{lost}$

$q_{lost}=-m\times c\times (T_2-T_1)$

where,

$q_{lost}$ = heat lost by the water = ?

c = specific heat of water = $4.184J/g^oC$

m = mass of water = 47.6 g

$T_1$ = initial temperature of water = $20.00^oC$

$T_2$ = final temperature of water = $11.00^oC$

Now put all the given values in the above formula, we get

$q_{lost}=-47.6g\times 4.184J/g^oC\times (11.00-20.00)^oC$

$q_{lost}=-1792.4J$

As,

$q_{absorbed}=-q_{lost}$

So,

$q_{absorbed}=1792.4J$

Therefore, the heat absorbed by the process is 1792.4 J