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LoloPo
An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10°C. Two metallic blocks are placed into the water. One is a 50.0 g piece of copper at 80.0°C; the other block has a mass of 70 g and is originally at a temperature of 100°C. The entire system stabilizes at a final temperature of 20°C. a) Determine the specific heat of the unknown sample. b) Determine a possible substance for the unknown material
The equation is Q cold = Q Hot 1. Find the specific heat of each Copper (Al) = 900 J/Kg*C Water (H2O) = 4186 J/Kg*C Copper (Cu) = 387 J/Kg*C unknown = x 2. Determine the energy gained or lost. In this problem the water and aluminum gain heat as they were at an equilibrium of 10 C but were raised to 20 C. The Copper was initially at 71 C and our unknown was at 100 C initially but they to ended up at 20 C, so they both lost energy. To do this we find each element separately by using the following formula : Q= MC (Tf-Ti) Q = energy lost or gained, M= mass of the element in kg, C is the specific heat of that element, Tf is the final temp and Ti is the initial temp of that element. Remember to convert from grams to Kg by moving decimal over 3 places to the left. Q (Al) = (.100 Kg) (900 J/Kg*C ) (20*C - 10*C) = 900 J/Kg*C (gained) Q (H2O) = (.250 Kg) (4186 J/Kg*C ) (20*C - 10*C) =10475 J/Kg*C (gained) Q (Cu) = (.05 Kg) (387 J/Kg*C ) (71*C - 20*C) = 987 J/Kg*C (Lost) Q (x) = (.084 Kg) ( x J/Kg*C ) (100*C - 20*C) = 6.72x J/Kg*C (Lost) 3. Now apply the formula (Q cold = Q Hot) by adding the numbers together. 900 J/Kg*C + 10475 J/Kg*C = 987 J/Kg*C + 6.72x J/Kg*C We have: 11375 J/Kg*C = 987 J/Kg*C + 6.72x J/Kg*C Subtract 987 from both sides: 11375 J/Kg*C = 987 J/Kg*C + 6.72x J/Kg*C -987 -987 10388 J/Kg*C = 6.72x J/Kg*C Divide by sides by 6.72 10388 J/Kg*C = 6.72x J/Kg*C /6.72 /6.72 X= 1545 J/Kg*C Thus the specific heat of our unknown substance is 1545 J/Kg*C
the first is unknown and i forgot the other