Answer:
22.44°C will be the final temperature of the water.
Explanation:
Heat lost by tin will be equal to heat gained by the water
[tex]-Q_1=Q_2[/tex]
Mass of tin = [tex]m_1=18.5 g[/tex]
Specific heat capacity of tin = [tex]c_1=0.21 J/g^oC [/tex]
Initial temperature of the tin = [tex]T_1=97.38^oC[/tex]
Final temperature = [tex]T_2[/tex]=T
[tex]Q_1=m_1c_1\times (T-T_1)[/tex]
Mass of water= [tex]m_2=75.7 g[/tex]
Specific heat capacity of water= [tex]c_2=4.184 J/g^oC [/tex]
Initial temperature of the water = [tex]T_3=21.52^oC[/tex]
Final temperature of water = [tex]T_2[/tex]=T
[tex]Q_2=m_2c_2\times (T-T_3)[/tex]
[tex]-Q_1=Q_2[/tex]
[tex]-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)[/tex]
On substituting all values:
[tex]-(18.5 g\times 0.21 J/g^oC\times (T-97.38^oC))=75.7 g\times 4.184 J/g^oC\times (T-21.52 ^oC)[/tex]
we get, T = 22.44°C
22.44°C will be the final temperature of the water.
Using the qunatity of heat relation, the final temperature of the water in degree Celcius ls 22.26°C
Recall :
Heat capacity of Tin = Heat gained by water
-mc△θ = mc△θ
Inputting the Parameters :
-(18.5 × 0.21 × (T - 97.38)) = (75.7 × 4.184 × (T - 21.52))
3.885(T - 97.38) = 316.7699(T - (21.52))
-3.885T + 378.3213 = 316.7288T - 6757.3574
Collect like terms :
378.3213 + 6757.3574 = 316.7288T + 3.885T
7135.6787 = 320.6138T
Divide both sides by 312.8849 isolate T
T = 20.26° C
Therefore, the temperature of the water is 22. 26°
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