Answer:
[tex]T_F=38.7\°C[/tex]
Explanation:
Hello,
In this case, considering that the copper is initially hot and the water is initially cold, we can infer that the heat lost by the copper is gained by the water as follows:
[tex]\Delta H_{Cu}=-\Delta _{w} H[/tex]
Which can be written in terms of temperatures, masses and heat capacities (390 mL equals 390 g for water).
[tex]m_{Cu}Cp_{Cu}(T_F-T_{Cu})=-m_{w}Cp_{w}(T_F-T_{w})[/tex]
In such a way, solving for the final temperature we obtain:
[tex]T_F=\frac{m_{Cu}Cp_{Cu}T_{Cu}+m_{w}Cp_{w}T_{w}}{m_{Cu}Cp_{Cu}+m_{w}Cp_{w}} \\\\T_F=\frac{248g*0.385J/(g\°C)*314\°C+390g*4.18J/(g\°C)*22.6\°C}{248g*0.385J/(g\°C)+390g*4.18J/(g\°C)} \\\\T_F=38.7\°C[/tex]
Regards.
The final temperature will be "38.71°C".
Let,
By applying the formula, we get
→ [tex]- (Specific \ heat\times Mass\times Temp \ Change) for \ copper = (Specific \ heat\times Mass\times Temp \ Change) for \ water[/tex]
By substituting the values, we get
→ [tex]-0.385\times 248\times (T-314) =4.184\times 390\times (T-22.6)[/tex]
→ [tex]-95.48(T-314) = 1631.76(T-22.6)[/tex]
→ [tex]-95.48 T+314\times 95.48=1631.76-22.6\times 1631.76[/tex]
→ [tex]1727.24 T=66858.496[/tex]
→ [tex]T = \frac{66858.496}{1727.24}[/tex]
[tex]= 38.71^{\circ} C[/tex]
Thus the above answer is correct.
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