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 84°C. The other sample has a mass of 58 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.
J/kg·°C
To determine the specific heat of the unknown second sample, we can use the principle of energy conservation. The heat lost by the hot sample (copper) and gained by the water and the calorimeter will be equal. We can calculate the quantity of heat exchanged for each substance involved using the formula:
Q = m * c * ΔT
Where:
Q is the heat exchanged
m is the mass of the substance
c is the specific heat of the substance
ΔT is the change in temperature
First, let's calculate the heat lost by the copper block. We know the initial temperature (T1), final temperature (T2), and mass (m) of the copper block. The specific heat of copper (c) is approximately 390 J/kg·°C.
Using the formula, we have:
Q1 = m * c * (T2 - T1)
Q1 = 50 g * 390 J/kg·°C * (20°C - 84°C)
Q1 = -169,800 J [Note: the negative sign indicates heat loss]
Next, let's calculate the heat gained by the water and the calorimeter. We know the mass of the water (m) and the change in temperature (ΔT). The specific heat of water (c) is approximately 4200 J/kg·°C.
Using the formula, we have:
Q2 = m * c * ΔT
Q2 = 250 g * 4200 J/kg·°C * (20°C - 10°C)
Q2 = 2,100,000 J
Since the heat lost by the copper block is equal to the heat gained by the water and the calorimeter, we can set up the equation:
Q1 = Q2
-169,800 J = 2,100,000 J
Now, let's solve for the specific heat (c) of the unknown second sample (let's denote it as c2).
Q1 = m2 * c2 * ΔT
-169,800 J = 58 g * c2 * (20°C - 100°C)
c2 = -169,800 J / (58 g * -80°C)
c2 ≈ 362 J/kg·°C
Therefore, the approximate specific heat of the unknown second sample is 362 J/kg·°C.