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39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of the two become equal. (a) What is the final temperature (in °C) when the two become equal? (Assume that coffee has the same specific heat as water, 4.1801 J/g·°C.) °C

Answer :

Answer: The final temperature of the solution is [tex]80.14^oC[/tex]

Explanation:

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, [tex]\text{heat}_{absorbed}=\text{heat}_{released}[/tex]

To calculate the amount of heat released or absorbed, we use the equation:  

[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]

Also,

[tex]m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)][/tex]    ..........(1)

where,

q = heat absorbed or released

[tex]m_1[/tex] = mass of aluminium = 39 g

[tex]m_2[/tex] = mass of coffee = 166 g

[tex]T_{final}[/tex] = final temperature = ?

[tex]T_1[/tex] = temperature of aluminium = [tex]24^oC[/tex]

[tex]T_2[/tex] = temperature of coffee = [tex]83^oC[/tex]

[tex]c_1[/tex] = specific heat of aluminium = [tex]0.904J/g^oC[/tex]

[tex]c_2[/tex] = specific heat of coffee= [tex]4.1801J/g^oC[/tex]

Putting all the values in equation 1, we get:

[tex]39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)][/tex]

[tex]T_{final}=80.14^oC[/tex]

Hence, the final temperature of the solution is [tex]80.14^oC[/tex]

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