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Question: What is the final temperature? \[{c_{Al}} = 0.215{\text{ cal }}{{\text{K}}^{{\text{ - 1}}}}{\text{...

What is the final temperature?
cAl=0.215 cal K - 1 g - 1{c_{Al}} = 0.215{\text{ cal }}{{\text{K}}^{{\text{ - 1}}}}{\text{ }}{{\text{g}}^{{\text{ - 1}}}}
25 g of Al → Ti  = 90 ∘C25{\text{ g of Al }} \to {\text{ }}{{\text{T}}_{{\text{i }}}}{\text{ = 90 }}^\circ {\text{C}}
In 100 g of water at Ti  = 20 ∘C{{\text{T}}_{{\text{i }}}}{\text{ = 20 }}^\circ {\text{C}}

Explanation

Solution

Assume that there is no heat loss from the system to the surrounding. The initial temperature of Al metal is greater than the initial temperature of the water. Heat lost by Al metal is equal to the heat gain by water.

Complete solution:
We have given the initial temperature and specific heat capacity of Al metal. Also, we have given an initial temperature of the water and we have to calculate the final temperature of the system
From the given initial temperatures of Al metal and water, we can say that the initial temperature of Al metal is greater than the initial temperature of the water. So, when we add Al metal to water, it will lose the heat while water will gain the same amount of heat energy.
So the relation between heat loss by Al metal and heat gain by water is as follows:
-Heat loss by Al metal = Heat gain water
The negative sign indicates a loss of heat.
The equation of heat, related to mass, specific heat capacity and the temperature difference is as follows:
q = msΔT{\text{q = ms}}\Delta T
Here,
q{\text{q}}= heat
m{\text{m}} = mass
s{\text{s}} = specific heat capacity
ΔT=Tf−Ti\Delta T = {T_f} - {T_i}
 - qAl=qH2O{\text{ - }}{{\text{q}}_{{\text{Al}}}} = {{\text{q}}_{{{\text{H}}_{\text{2}}}{\text{O}}}}
We can rewrite this equation as follows:
( - msΔT)Al=(msΔT)H2O{{\text{( - ms}}\Delta T)_{{\text{Al}}}} = {({\text{ms}}\Delta T)_{{{\text{H}}_{\text{2}}}{\text{O}}}}
Specific heat capacity of water is 1 cal K - 1 g - 11{\text{ cal }}{{\text{K}}^{{\text{ - 1}}}}{\text{ }}{{\text{g}}^{{\text{ - 1}}}}
Now, by substituting all given data we can calculate the final temperature as follows:
−[25 g×0.215 cal K - 1 g - 1(Tf− 90 ∘C)]=100 g ×1 cal K - 1 g - 1(Tf− 20 ∘C)- [25{\text{ g}} \times 0.215{\text{ cal }}{{\text{K}}^{{\text{ - 1}}}}{\text{ }}{{\text{g}}^{{\text{ - 1}}}}({T_f} - {\text{ 90 }}^\circ {\text{C}})] = 100{\text{ g }} \times 1{\text{ cal }}{{\text{K}}^{{\text{ - 1}}}}{\text{ }}{{\text{g}}^{{\text{ - 1}}}}({T_f} - {\text{ 20 }}^\circ {\text{C}})
−5.375 - 1(Tf− 90 ∘C)=100(Tf− 20 ∘C)- {5.375^{{\text{ - 1}}}}({T_f} - {\text{ 90 }}^\circ {\text{C}}) = 100({T_f} - {\text{ 20 }}^\circ {\text{C}})
−0.05375(Tf− 90 ∘C)=(Tf− 20 ∘C)- 0.05375({T_f} - {\text{ 90 }}^\circ {\text{C}}) = ({T_f} - {\text{ 20 }}^\circ {\text{C}})
−0.05375Tf+4.8375=Tf− 20 - 0.05375{T_f} + 4.8375 = {T_f} - {\text{ 20 }}
1.05375Tf = 24.8375{\text{1}}{\text{.05375}}{T_f}{\text{ = 24}}{\text{.8375}}
Tf=23.57∘C{T_f} = 23.57^\circ {\text{C}}

Hence, the final temperature is 23.57∘C23.57^\circ {\text{C}}.

Note: Specific heat capacity is the amount of heat required to raise the temperature of 1g of a substance by 1∘C{\text{1}}^\circ {\text{C}}. A system that gains the heat has a final temperature greater than the initial temperature. While a system that losses heat has a final temperature less than the initial temperature.