Question

A 13.60-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_____ °C

Answer 1

Answer : The final temperature of the aluminum and ethylene glycol is
`56.2^oC`

Explanation :

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

`q_1=-q_2`

`m_1* c_1* (T_f-T_1)=-m_2* c_2* (T_f-T_2)`

where,

`c_1` = specific heat of aluminum =
`0.900J/g^oC`

`c_2` = specific heat of ethylene glycol =
`2.36J/g^oC`

`m_1` = mass of aluminum = 13.60 g

`m_2` = mass of ethylene glycol = 23.27 g

`T_f` = final temperature of aluminum and ethylene glycol = ?

`T_1` = initial temperature of aluminium =
`13.91^oC`

`T_2` = initial temperature of ethylene glycol =
`65.66^oC`

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

`13.60g* 0.900J/g^oC* (T_f-13.91)^oC=-23.27g* 2.36J/g^oC* (T_f-65.66)^oC`

`T_f=56.2^oC`

Therefore, the final temperature of the aluminum and ethylene glycol is
`56.2^oC`