Respuesta :
Answer:
The specific heat capacity of the unknown metal is C = 0.6991 J/g°C = 0.1671 cal/g°C
Explanation:
Heat lost by the unknown metal is equal to the heat gained by the water and aluminium cup.
Given,
Mass of unknown metal = 77 g
Initial Temperature of unknown metal = 99°C
Mass of water = 225 g
Initial Temperature of water = 22°C
Mass of Aluminium cup = 44 g
Specific heat capacity of Aluminium cup = 0.22 cal/gᵒC = 0.92048 J/g°C
Final temperature of the setup = 26°C
Note that, specific heat capacity of water = 4.186 J/g°C
Let the specific heat of the unknown metal be C
Heat lost from the unknown metal
= (77)(C)(99 - 26) = (5,621C) J
Heat gained by water
= (225)(4.186)(26 - 22) = 3,767.4 J
Heat gained by Aluminium cup
= (44)(0.92048)(26 - 22) = 162.00448 J
Heat lost by unknown metal = (Heat gained by water) + (Heat gained by Aluminium cup)
5621C = 3,767.4 + 162.00448 = 3,929.40448
5621C = 3,929.40448
C = (3,929.40448 ÷ 5621) = 0.6991 J/g°C
C = 0.6991 J/g°C = 0.1671 cal/g°C
Hope this Helps!!!
Answer:
[tex]Cp_{metal}=0.17\frac{cal}{g^oC}[/tex]
Explanation:
Hello,
In this case, for the given equilibrium, the following equality is accomplished for the involved energy transfer in the system:
[tex]\Delta H_{metal}+\Delta H_{water}+\Delta H_{Al\ cup}=0[/tex]
Thus, in terms of masses, heat capacities and temperatures:
[tex]m_{metal}Cp_{metal}(T_{eq}-T_{metal})+m_{Al\ cup}Cp_{Al\ cup}(T_{eq}-T_{Al\ cup})+m_{water}Cp_{water}(T_{eq}-T_{water})=0[/tex]
Hence, solving for the heat capacity of the metal:
[tex]Cp_{metal}=\frac{-m_{Al\ cup}Cp_{Al\ cup}(T_{eq}-T_{Al\ cup})-m_{water}Cp_{water}(T_{eq}-T_{water})}{m_{metal}(T_{eq}-T_{metal})} \\\\Cp_{metal}=\frac{-44g*0.22\frac{cal}{g^oC} *(26-22)^oC-225g*1\frac{cal}{g^oC}*(26-22)^oC}{77g(26-99)^oC} \\\\Cp_{metal}=0.17\frac{cal}{g^oC}[/tex]
Best regards.
