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

c. heat could easily travel through the molecules in the air and escape/enter the flask. removing the air slows down the heat transfer.  

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