In an experiment,30.0 g of metal is heated to 98.0°C and then quickly transferred to 50.0 g of water in a calorimeter at 27.0°C.The heat capacity of the calorimeter with the water is 211 J/°C.The final temperature comes to 32.5°C.What is the approximate specific heat capacity of the metal? [c_p(water)= 4.18 J/(g.°C)]

Question

Quizplus · Accepted Answer

First, we need to calculate the heat absorbed by the water in the calorimeter:
qwater = mwater x cwater x ΔT
qwater = 50.0 g x 4.18 J/(g°C) x (32.5°C - 27.0°C)
qwater = 1171 J

Since the metal is the only other object involved in the process, the heat lost by the metal is equal to the heat gained by the water:
qmetal = -qwater
qmetal = -1171 J

Now, we can use the specific heat capacity equation to calculate the specific heat capacity of the metal:
qmetal = mm x cmetal x ΔTmetal
-1171 J = 30.0 g x cm x (32.5°C - 98.0°C)
cm = 0.591 J/(g°C)

Therefore, the answer is D, approximately 0.591 J/(g°C).

In an Experiment,30