In the laboratory a "coffee cup" calorimeter, or constant pressure calorimeter, is frequently used to determine the specific heat of a solid, or to measure the energy of a solution phase reaction. a student heats 61.68 grams of gold to 99.01 °c and then drops it into a cup containing 79.34 grams of water at 22.14 °c. she measures the final temperature to be 23.98 °c. the heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.80 j/°c. assuming that no heat is lost to the surroundings calculate the specific heat of gold.

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:

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

b is the correct answer

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