Specific Heat in cal/g*degreesC

Question

A metal object with mass 27.5g is heated to 97.0 degrees Celsius and then transferred to an insulated container containing 84.4g of water at 20.5 degrees Celsius.  The water temperature rises and the temperature of the metal object falls until they both reach the same final temperature of 24.9 degrees Celsius.  What is the specific heat of this metal object? Assume that all the heat lost by the metal object is absorbed by the water.

Answer 1

Heat gained by the water = mcT , where m is the mass of water , c is specific heat capacity of water and T is the Temperature change

Q= 84.4 g * 4.4 C * 1 cal / g C = 371.36 calorie

we assume that the heat gained by water is equal to heat lost by the metal , i.e. the metal lost 371.36 calorie of heat.

temperature change by the metal = 97.0 - 24.9 = 72.1 C

let the heat capacity of metal be C₁

_{heat lost = mass * heat capacity * temp change = 371.36 = 27.5 * 72.1 * C1}

_{C1= 0.187 cal / g C}

_{so the specific heat capacity of the metal is 0.187 cal / g C ( 3 significant figures )}

Answer 2

heat = mC∆T where m = mass; C = specific heat; ∆T = change in temperature and q = heat

For the metal:

q = mC∆T = (27.5 g)(C)(97º - 24.9º) = (27.5)(C)(69.5 deg)

For the water:

q = mC∆T = (84.4 g)(1 cal/g/deg)(24.9º - 20.5º) = (84.4)(4.4 deg) = 371 cal of heat

The heat lost by the metal must equal the heat gained by the water, so we can write...

371 cal = (27.5 g)(C)(69.5 deg)

C = 371 cal/(27.5 g)(69.5 deg)

C = 0.194 cal/g/deg = specific heat of the metal