Answer:
The specific heat of the metal is 0.0482 joules per gram-Celsius.
Explanation:
Block of metal is cooled by adding water and thermal equilibrium is reached. According to the First Law of Thermodynamics and by supposing the absence of energy and mass interaction of the system with surroundings, the change in the energy of the system is represented by:
[tex]\Delta U_{m} + \Delta U_{w} = 0[/tex]
Where [tex]\Delta U_{m}[/tex] and [tex]\Delta U_{w}[/tex] are the changes in internal energies of the block of metal and water, measured in joules.
The expression described above is now extended by applying the definition of internal energy for constant mass systems:
[tex]m_{m}\cdot c_{v,m}\cdot (T_{m,in}-T_{m,out}) + m_{w}\cdot c_{v,w}\cdot (T_{w,in}-T_{w,out})= 0[/tex]
Where:
[tex]m_{m}[/tex], [tex]m_{w}[/tex] - Masses of the block of metal and water, measured in grams.
[tex]c_{v,m}[/tex], [tex]c_{v,w}[/tex] - Specific heats of the block of metal and water, measured in joules per gram-Celsius.
[tex]T_{in, m}[/tex], [tex]T_{in, w}[/tex] - Initial temperatures of the block of metal and water, measured in Celsius.
[tex]T_{out, m}[/tex], [tex]T_{out, w}[/tex] - Final temperatures of the block of metal and water, measured in Celsius.
The specific heat of the metal is cleared in the equation:
[tex]c_{v,m} = -\frac{m_{w}\cdot c_{v,w} (T_{w,in}-T_{w,out})}{m_{m}\cdot (T_{m,in}-T_{m,out})}[/tex]
If [tex]m_{m} = 12.6\,g[/tex], [tex]m_{w} = 100\,g[/tex], [tex]c_{v,w} = 4,186\,\frac{J}{kg\cdot ^{\circ}C}[/tex], [tex]T_{w,in} = 21\,^{\circ}C[/tex], [tex]T_{m,in} = 94\,^{\circ}C[/tex] and [tex]T_{m,out} = T_{w,out} = 25.1\,^{\circ}C[/tex], the specific heat of the metal is:
[tex]c_{v,m} = -\frac{(100\,g)\cdot \left(4,186\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (25\,^{\circ}C-25.1\,^{\circ}C)}{(12.6\,g)\cdot (94\,^{\circ}C-25.1\,^{\circ}C)}[/tex]
[tex]c_{v,m} = 0.0482\,\frac{J}{kg\cdot ^{\circ}C}[/tex]
The specific heat of the metal is 0.0482 joules per gram-Celsius.
