In an experiment, 138 g of graphite is heated to 90.0C and then quickly transferred to 150.0 g of water in a calorimeter at 25.011eed8b2_2d82_a951_a0ba_55ddc23a4114_TB6562_11C.The heat capacity of the calorimeter with the water is 157 J/11eed8b2_2d82_a951_a0ba_55ddc23a4114_TB6562_11C.The final temperature comes to 50.011eed8b2_2d82_a951_a0ba_55ddc23a4114_TB6562_11C.What is the approximate specific heat capacity of graphite?
[c_p(water) = 4.18 J/(g .11eed8b2_2d82_a951_a0ba_55ddc23a4114_TB6562_11C) ]

Question

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 25.011eed8b2_2d82_a951_a0ba_55ddc23a4114_TB6562_11C.The heat capacity of the calorimeter with the water is 157 J/11eed8b2_2d82_a951_a0ba_55ddc23a4114_TB6562_11C.The final temperature comes to 50.011eed8b2_2d82_a951_a0ba_55ddc23a4114_TB6562_11C.What is the approximate specific heat capacity of graphite?
[c_p(water) = 4.18 J/(g .11eed8b2_2d82_a951_a0ba_55ddc23a4114_TB6562_11C) ]

Quizplus · Accepted Answer

The heat gained by the water and calorimeter is equal to the heat lost by the graphite. Using the formula $q = mc\Delta T$, where $q$ is the heat transferred, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature, we can set up the equation for the water and calorimeter gaining heat equal to the graphite losing heat. For water and calorimeter: $q_{water} = (150.0\,g)(4.18\,J/g^\circ C)(50.0^\circ C - 25.0^\circ C) + (157\,J/^\circ C)(50.0^\circ C - 25.0^\circ C)$. For graphite: $q_{graphite} = (138\,g)(c_{graphite})(90.0^\circ C - 50.0^\circ C)$. Solving for $c_{graphite}$ gives approximately $0.711\,J/(g^\circ C)$.

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