Final answer:
The question involves using the specific heat equation to determine the final temperature when silver and water are mixed without heat loss to the surroundings. One needs to set the heat lost by the silver equal to the heat gained by the water and solve for final temperature.
Step-by-step explanation:
The question involves the application of the specific heat equation in thermodynamics, which relates the mass of a substance, its specific heat, and change in temperature to the heat exchanged. For a system with no heat loss to the surroundings, the heat lost by the silver must equal the heat gained by the water. The specific heat of silver is provided as 0.0565 cal/g°C, and the specific heat of water is 1.0 cal/g°C.
The equation used is q = m × c × ΔT, where q is heat in calories, m is mass in grams, c is specific heat in cal/g°C, and ΔT is the change in temperature in degrees Celsius. For the silver (Ag), qAg = mAg × cAg × (ΔTAg), and for the water (H2O), qH2O = mH2O × cH2O × (ΔTH2O). Since the system is isolated, we have the heat gained by water equals the heat lost by silver: qAg = -qH2O.
The temperatures can be represented as Tfinal - Tinitial, yielding the equation for equilibrium: mAg × cAg × (Tfinal - Tinitial, Ag) = - (mH2O × cH2O × (Tfinal - Tinitial, H2O)).
To solve for the final temperature, the equation is rearranged, allowing for the cancellation of q's negative sign and prompting a final equation that can be solved for Tfinal using algebra. This provides the final temperature reached by both the silver and the water.