Acoffee cup calorimeter contains 27.100 g water at 25.0°c. a 182.000 g sample of gold (specific heat capacity of 0.128 j/gºc) is added to the coffee cup calorimeter. the final
temperature is 27.5°c. what was the initial temperature of the gold?

15.33° C

We are given the following;

Mass of water as 27.10 g

Initial temperature of water is 25°C

But, K = °C + 273.15

Thus, temperature of water is 298.15 K

Specific heat capacity of gold is 0.128 J/g°C

Mass of gold sample as 182.0 g

The final temperature of the mixture is 27.5°C

we are required to calculate the initial temperature of gold sample;

We know that, heat absorbed, Q = mcΔT

Change in temperature, ΔT = 27.5°C - 25°C

                                              = 2.5°C

Therefore;

Heat absorbed by water = 27.10 g × 4.184 J/g°C × 2.5°C

                                         = 283.466 Joules

Q = mcΔT

Assuming the initial temperature of gold sample is X°C

Therefore, change in temperature, ΔT = (27.5 - X)°

Thus;

Q = 182.0 g × 0.128 J/g°C × (27.5 - X)°

  = 640.64 - 23.296X Joules

We need to know that;

Heat absorbed by water is equivalent to heat released by the gold sample

Therefore;

640.64 - 23.296X Joules = 283.466 Joules

Combining the like terms ;

23.296X = 357.174

           X = 15.33°C

Therefore, the initial temperature of the gold sample is 15.33°C

fe 55.85 g/mol and h2o 18.02 g/mol.