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A metal weighing 4.82 g was heated to 115.0 °C and put into 35 mL of water of temperature 28.7 °C. The metal and water were allowed to come to an equilibrium temperature, determined to be 34.5 °C. Assuming no heat was lost to the environment, calculate the specific heat of the metal. Consider the specific heat capacity of water as 4.186 joule/gram °C.The heat absorbed by the water is J (whole #)What is the specific heat of the metal? J/gC (tenth)

Respuesta :

We will have the following:

At equilibrium the specific heats will be:

[tex]m_{metal}\ast c_{metal}\ast\Delta t=m_{h20}\ast c_{h2o}\ast\Delta t[/tex]

So:

[tex]\begin{gathered} (4.82g)(c)(34.5C-115.0C)=(35g)(4.186J/g\ast C)(34.5C-28.7C) \\ \\ \Rightarrow c=\frac{(35g)(4.186J/g\ast C)(34.5C-28.7C)}{(4.82g)(34.5C-115.0C)}\Rightarrow c=-\frac{2639}{1205}J/g\ast C \\ \\ \Rightarrow c\approx-2.2J/g\ast C \end{gathered}[/tex]

So, the specific heat of the metal will be approximately 2.2 J/g*°C.

*The heat absorbed by the water will be:

[tex]Q=(25g)(4.186J/g\ast C)(34.5C-28.7C)\Rightarrow Q=606.97J[/tex]

So, the water absorbed 606.97 Joules.

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