Answer to Question #88291 in General Chemistry for tes

Question

A foundry worker places a 5.78 kg sheet of cobalt at a temperature of 15°C on top of a 16.6 kg sheet of palladium at 68°C. Assuming no heat is lost to the surroundings, calculate the final temperature of the two sheets of metal.

Answer 1

ΔQ = mcΔT

ΔQ - the heat energy put into or taken out of the substance,

m - the mass of the substance,

c - the specific heat capacity,

ΔT - the temperature differential where the initial temperature of the reaction is subtracted from the final temperature. 

Since no heat is lost the heat energy that is gained by the cobalt is equal to the heat energy that is given off by the palladium

c(cobalt) = 0.41868 J/(g×°C)

c(palladium) = 0.23864 J/(g×°C)

ΔQ(cobalt) = 5.78×10³ g x 0.41868 J/(g×°C) x (T_f – 15 °C) = 2419.9704 T_f – 36299.556

ΔQ(palladium) = 16.6×10³ g x 0.23864 J/(g×°C) x (68 °C – T_f) = 269376.832 – 3961.424 T_f 

ΔQ(cobalt) = ΔQ(palladium)

2419.9704 T_f – 36299.556 = 269376.832 – 3961.424 T_f

6381.3944 T_f = 305676.388

T_f = 47.9 °C