Question

75.0 gram sample of silver at 100. °C is added to a calorimeter containing 100. G of water at 20.0°C. Given that the specific heat capacity of silver is 0.235 J/g°C and the specific heat of water is 4.184 J/g°C, what is the final temperature of the water/silver mixture when thermal equilibrium is achieved?

Answer 1

The final temperature of a water/silver mixture at thermal equilibrium is determined using the conservation of energy equation. By setting the heat lost by silver equal to the heat gained by water and solving for the final temperature, one can find the equilibrium temperature of the mixture.

Step-by-step explanation:

The final temperature of a water/silver mixture when thermal equilibrium is achieved can be determined using the principle of conservation of energy. In this scenario, energy lost by the hot silver must equal the energy gained by the cooler water until both reach the same final temperature. To find this temperature, we use the equation Q = mcΔT, where Q is heat energy, m is mass, c is specific heat capacity, and ΔT is the change in temperature.

We set the heat lost by silver equal to the heat gained by water:

(75.0 g)(0.235 J/g°C)(Tfinal - 100 °C) = (100.0 g)(4.184 J/g°C)(Tfinal - 20.0 °C)

This equation can be solved for Tfinal which would provide the equilibrium temperature of the mixture. Step by step, you would distribute the mass and specific heat through the parentheses, combine like terms, and isolate Tfinal to solve for the final temperature.