A13.60-g block of solid aluminum at 13.91 °c is immersed in a 23.27-g pool of liquid ethylene glycol with a temperature of 65.66 °c. when thermal equilibrium is reached, what is the temperature of the aluminum and ethylene glycol? specific heat capacities: lead = 0.159 j/g °c; ethylene glycol = 2.36 j/g ° °c

Answer : The final temperature of the aluminum and ethylene glycol is

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

where,

= specific heat of aluminum =

= specific heat of ethylene glycol =

= mass of aluminum = 13.60 g

= mass of ethylene glycol = 23.27 g

= final temperature of aluminum and ethylene glycol = ?

= initial temperature of aluminium =

= initial temperature of ethylene glycol =

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

Therefore, the final temperature of the aluminum and ethylene glycol is

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

it is the most commonly accepted one