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poizon [28]
4 years ago
6

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 fina

l 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
Chemistry
1 answer:
Darina [25.2K]4 years ago
7 0

<u>Answer:</u> The final temperature will be 23^oC

<u>Explanation:</u>

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

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