A 1.00 x 102-g block of unknown material at 100.0°C is placed in 1.00 x 102g of water at 10.0°C. The final temperature of the mixture is 25.0°C.

Question

A 1.00 x 102-g block of unknown material at 100.0°C is placed in 1.00 x 102g of water at 10.0°C. The final temperature of the mixture is 25.0°C. What is the specific heat of the block? Which material might the block be made of?

Please show work!

Answer 1

m₁ = mass of block (given in problem)

T₁ = initial temperature of block (given in problem)

c₁ = specific heat of block = ?

m₂ = mass of water (given in problem)

T₂ = initial temperature of water (given in problem)

c₂ = specific heat of water (look it up)

T = final equilibrium temperature (given in problem)

Use conservation of energy:

m₁c₁(T₁ - T) = m₂c₂(T - T₂)

Plug in the numbers and solve for c₁. Be careful with units. Then go to a table of specific heats and find which material has the same specific heat you just found.