Answer:
Specific heat of gold is 0,133J/g°C
Explanation:
In this problem, the heat of the gold is transferred to water and the calorimeter, that means:
[tex]q_{Lost By Metal} = q_{GainedWater} + q_{Gained Calorimeter}[/tex]
The Q lost by metal is:
Q = C×m×ΔT, Where m is mass (61,68g), ΔT is (99,01°C-23,98°C = 75,03°C) and C is sepecific heat of gold
The Q gained by water is:
Q = C×m×ΔT, Where m is mass (79,34g), ΔT is (23,98°C-22,14°C = 1,84°C) and C is sepecific heat of water (4,184J/g°C)
The Q gained by calorimeter is:
Q =Cc×ΔT Where Cc is calorimeter constant (1,80J/°C), and ΔT is (23,98°C-22,14°C = 1,84°C)
Replacing:
C×61,68g×75,03°C = 4,184J/g°C×79,34g×1,84°C + 1,80J/°C×1,84°C
4628g°C×C = 610,8J + 3,3J
4628g°C×C = 614,1J
C = 0,133J/g°C
I hope it helps
