Question

An unknown piece of metal weighing 123 g is placed in boiling water and after a few minutes is plunged into 75 mL of water at 25 degrees Celsius. If the final temperature of the metal and water is 29 degrees Celsius, what is the specific heat capacity of the metal? (Assume that the density of water is 1.0 g/mL and specific heat capacity of water is 4.184 J/g degrees Celsius).

Answer 1

The specific heat capacity of the metal is 0.75 J/g°C.

Step-by-step explanation:

To find the specific heat capacity of the metal, we can use the equation: Q = mcΔT, where Q is the heat absorbed or released, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. We can calculate the heat absorbed by the metal using the equation Q = mcΔT, where m is the mass of the metal and ΔT is the change in temperature. First, let's calculate the heat absorbed by the metal: Q = (123 g)(4.184 J/g°C)(29°C - 25°C) = 2043 J. Now, let's calculate the heat absorbed by the water: Q = (75 mL)(1.0 g/mL)(4.184 J/g°C)(29°C - 25°C) = 924 J.

Since the metal and water reach the same final temperature, the heat gained by the metal is equal to the heat lost by the water. Therefore, 2043 J = 924 J. Rearranging the equation to solve for the specific heat capacity of the metal, we get: c = Q / (mΔT) = 924 J / (123 g)(29°C - 25°C) = 0.75 J/g°C. Therefore, the specific heat capacity of the metal is 0.75 J/g°C.