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 $23^oC$

Explanation:

Calculating the heat released or absorbed for the process:

$q=m\times C\times (T_2-T_1)$

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

$q_1=-q_2$

    OR

$m_1\times C_1\times (T_f-T_1)=-m_2\times C_2\times (T_f-T_2)$           ......(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:

$30\times 0.21\times (T_f-40)=-50\times 1\times (T_f-21)\\\\56.3T_f=1302\\\\T_f=\frac{1302}{56.3}=23.13^oC=23^oC$

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