77 grams of an unknown metal at 99ᵒC is placed in 225 grams of water which is initially at 22ᵒC. The water is inside a 44 gram a

Question

3 years ago

8

77 grams of an unknown metal at 99ᵒC is placed in 225 grams of water which is initially at 22ᵒC. The water is inside a 44 gram a

luminum cup with a specific heat of 0.22 cal/gᵒC. The final temperature of the system is 26ᵒC. What is the specific heat of the unknown metal?

Chemistry

2 answers:

KengaRu [80] · Answer 1 · 2022-07-20T17:06:29+03:00

KengaRu [80]3 years ago

7 0

Answer:

$Cp_{metal}=0.17\frac{cal}{g^oC}$

Explanation:

Hello,

In this case, for the given equilibrium, the following equality is accomplished for the involved energy transfer in the system:

$\Delta H_{metal}+\Delta H_{water}+\Delta H_{Al\ cup}=0$

Thus, in terms of masses, heat capacities and temperatures:

$m_{metal}Cp_{metal}(T_{eq}-T_{metal})+m_{Al\ cup}Cp_{Al\ cup}(T_{eq}-T_{Al\ cup})+m_{water}Cp_{water}(T_{eq}-T_{water})=0$

Hence, solving for the heat capacity of the metal:

$Cp_{metal}=\frac{-m_{Al\ cup}Cp_{Al\ cup}(T_{eq}-T_{Al\ cup})-m_{water}Cp_{water}(T_{eq}-T_{water})}{m_{metal}(T_{eq}-T_{metal})} \\\\Cp_{metal}=\frac{-44g*0.22\frac{cal}{g^oC} *(26-22)^oC-225g*1\frac{cal}{g^oC}*(26-22)^oC}{77g(26-99)^oC} \\\\Cp_{metal}=0.17\frac{cal}{g^oC}$