Question

A 30. g sample of Aluminum was heated to 40. 0C and placed in a calorimeter containing 50. g of water at 21 0C. What is the final temperature of the aluminum-water system if the cAl = 0.21 cal/g0C and cwater = 1.0 cal/ g 0C.

Write the complete equation you will use. 1 point

Substitute the values in the equation in step 1 . 1 point

Report the math answer with 2 sig figs and the correct unit. 1 point

Answer 1

Answer: The final temperature will be
`23^oC`

Step-by-step explanation:

Calculating the heat released or absorbed for the process:

`q=m* C* (T_2-T_1)`

In a system, the total amount of heat released is equal to the total amount of heat absorbed.

`q_1=-q_2`

OR

`m_1* C_1* (T_f-T_1)=-m_2* C_2* (T_f-T_2)` ......(1)

where,

`C_1` = specific heat of aluminium =
`0.21 Cal/g^oC`

`C_2` = heat capacity of water =
`1Cal/g^oC`

`m_1` = mass of aluminium = 30. g

`m_2` = mass of water = 50. g

`T_f` = final temperature of the system = ?

`T_1` = initial temperature of aluminium =
`40.^oC`

`T_2` = initial temperature of the water =
`21.^oC`

Putting values in equation 1, we get:

`30* 0.21* (T_f-40)=-50* 1* (T_f-21)\\\\56.3T_f=1302\\\\T_f=(1302)/(56.3)=23.13^oC=23^oC`

Hence, the final temperature will be
`23^oC`