Final answer:
By using the heat transfer formula and converting the given heat capacities into consistent units, we find that after adding 2.00 kJ of heat, the final temperature of the silver slug is 40.27°C and of the lead slug is 27.48°C.
Step-by-step explanation:
To determine the final temperature of each slug when 2.00 kJ of heat is added to a slug of silver and a slug of lead, we can use the formula q = mcΔT, where q is the heat transfer, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. However, the mass (m) of the slugs is not provided here, so we will convert the specific heat capacities given in J/°C to kJ/°C to match the unit of heat added. By manipulating the formula as ΔT = q/(mc), we can find ΔT for both metals using the provided heat capacities.
For the silver slug, we get ΔT = 2.00 kJ / (131 J/°C • 1 kJ/1000 J) = 15.27°C. Hence, the final temperature of the silver is 25.00°C + 15.27°C = 40.27°C.
For the lead slug, ΔT = 2.00 kJ / (805 J/°C • 1 kJ/1000 J) = 2.48°C. Thus, the final temperature of the lead is 25.00°C + 2.48°C = 27.48°C.