Chem: Calorimetry

Question

A 25.00 gram sample of an unknown metal initially at 99.0 degrees Celcius is added to 50.00 grams of water initially at 6.00 degrees Celcius. The final temperature of the system is 20.15 degrees Celcius. Calculate the specific heat of the metal.  J/gC

Answer 1

Heat from the metal @ 99º will be transferred to the cooler water @ 6.00º.

Heat LOST by metal MUST EQUAL heat GAINED by the water. This is conservation of energy.

heat lost by metal = q = mC∆T

m = mass of metal = 25.00 g

C = specific heat of metal = ?

∆T = change in temperature of metal = 99.0º - 20.15º = 78.85º

q = (25.00 g)(C)(78.85º)

heat gained by water = q = mC∆T

m = mass of water = 50.00 g

C = specific heat of water = 4.184 J/gº

∆T = change in temperature of water = 20.15º - 6.00º = 14.15º

q = (50.00 g)(4.184 J/gº)(14.15º) = 2960.2 J

So, we now know that 2960.2 J of energy were lost by the metal. Use this to solve for C of metal:

For the metal, we have

q = (25.00 g)(C)(78.85º)

2960.2 J = (25.00 g)(C)(78.85º )

2960.2 J / (25.00 g)(78.85ª) = C

C_metal = 1.50 J/gº (3 sig. figs.)