A 100 g aluminum calorimeter contains 250 g of water. The two substances are in thermal equilibrium at 10°C. Two metallic blocks are placed in the water. One is a 50 g piece of copper at 66°C. The other sample has a mass of 78 g and is originally at a temperature of 100°C. The entire system stabilizes at a final temperature of 20°C. Determine the specific heat of the unknown second sample

### 4 Answers

Specific heat of aluminum = 0.897 J / g degC.

Heat gained by calorimeter = 100 * 10 * 0.897 J = 897 J

Specific heat of water = 4.18 J / g degC.

Heat gained by water = 250 * 10 * 4.18 J = 10450 J

Specific heat of copper = 0.385 J / g degC.

Heat loss by copper block = 50 * (66-10) * 0.385 J = 1078 J

Let specific heat of unknown block be x J / g degC.

Heat loss by unknown block = 78 * (100-10) * x J = 7020x J

Heat gain = Heat loss

897 + 10450 = 1078 + 7020x

7020x = 10269

x =~ 1.463

Specific heat of unknown substance is 1.463 J / g degC.

We have several components to keep track of in this system so let’s list them out. 250 g water at 10 C 100 g aluminum at 10 C 50 g copper at 83 C 70 g unknown at 100 C To attain thermal equilibrium, water and aluminum must gain heat and copper and the unknown must lose heat. The sum of the heats gained must equal the sum of the heats lost. Write this out mathematically. Q(gained aluminum) + Q(gained water) = -(Q(lost copper) + Q(lost unknown)) Q = cmΔT, where c is the specific heat capacity, m is the mass of substance, and ΔT is the final minus initial temperature. Aluminum: cmΔT = cm(20-10) = 10cm Water: cmΔT = 10cm Copper: cmΔT = cm(20-83) = -63cm Unknown: cmΔT = cm(20-100) = -80cm The “cm” terms are the product of the specific heat of the respective substance and the mass of the substance. The “c” value varies with the substance and the masses are given. The only unknown at this point is the “c” value for the unknown, which can be readily calculated using some algebra. Hope this helps.

Aluminum Calorimeter

answer is wrong…