Respuesta :
Answer:
The specific heat of the metal is 0.69 J/g°C.
Explanation:
The given information from the exercise is:
• Metal,:
- Mass (massmetal): 52.3g
- Initial temperature (Tinitialmetal): 126.2°C
- Final temperature (Tfinalmetal): 35.5°C
• Water,:
- Mass (masswater):71.2g
- Initial temperature (Tinitialwater): 24.5°C
- Final temperature (Tfinalwater): 35.5°C
1st) It is necesary to write the Heat formula for both materials, the metal and water. In the case of water, we can solve the formula, because we have all the nedeed information:
• Metal,:
[tex]\begin{gathered} Q_m=m_{metal}*c_{metal}*(T_{fmetal}-T_{imetal}) \\ Q_m=52.3g*c_{metal}*(35.5_\degree C-126.2\degree C) \\ Q_m=-4,743.61g\degree C*c_{metal} \\ \end{gathered}[/tex]Water:
[tex]\begin{gathered} Q_w=m_{water}*c_{water}*(T_{fwater}-T_{\imaginaryI water}) \\ Q_w=71.2g*4.184\frac{J}{g\degree C}*(35.5\degree C-24.5\degree C) \\ Q_w=3,276.9J \end{gathered}[/tex]So, the heat absorbed by the water is 3,276.9 J.
2nd) In the end both materials reach the equilibrium temperature, because the heat released by the piece of metal is absorbed by the water, this is represented as:
[tex]-Q_m=Q_w[/tex]Finally, we have to replace the heat of the metal equation (Qm) and the heat of water (Qw) to calculate the specific heat of the metal:
[tex]\begin{gathered} -Q_{m}=Q_{w} \\ -(-4,743.61g)\degree C*c_{metal}=3,276.9J \\ 4,743.61g\degree C*c_{metal}=3,276.9J \\ c_{metal}=\frac{3,276.9J}{4,743.61g\degree C} \\ c_{metal}=0.69\frac{J}{g\degree C} \end{gathered}[/tex]So, the specific heat of the metal is 0.69 J/g°C.
