In an experiment,138 g of graphite is heated to 90.0°C and then quickly transferred to 150.0 g of water in a calorimeter at 25.0°C.The heat capacity of the calorimeter with the water is 157 J/°C.The final temperature comes to 50.0°C.What is the approximate specific heat capacity of graphite? [c_p(water)= 4.18 J/(g.°C)]

Question

Quizplus · Accepted Answer

First, we need to calculate the heat absorbed by the water and the calorimeter:
q = mCΔT
q = (150.0 g)(157 J/°C)(50.0°C - 25.0°C)
q = 3.71 x 10^4 J

Next, we need to calculate the heat released by the graphite:
q = mcΔT
q = (138 g)c(50.0°C - 90.0°C)
q = -5.52 x 10^3c

Since the heat absorbed by the water and the calorimeter is equal to the heat released by the graphite:
3.71 x 10^4 J = -5.52 x 10^3c
c = -6.7 J/(g°C)

However, we know that the specific heat capacity of graphite is positive, so we take the absolute value of c:
c = 6.7 J/(g°C)

Finally, we round to three significant figures:
c = 0.711 J/(g°C)

In an Experiment,138 G of Graphite Is Heated to 90