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 10.67 degrees Celcius. The final temperature of the system is 20.15 degrees Celcius. Calculate the specific heat of the metal. (The specific heat of water is 4.184 J/g*C).
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 10.67 degrees Celcius. The final temperature of the system is 20.15 degrees Celcius. Calculate the specific heat of the metal. (The specific heat of water is 4.184 J/g*C).
Solution
To solve this problem, we will use the principle of conservation of energy, which states that the heat lost by the hot object (the metal) will be equal to the heat gained by the cold object (the water).
The formula for heat transfer is Q = mcΔT, where:
- Q is the heat transferred,
- m is the mass of the substance,
- c is the specific heat of the substance, and
- ΔT is the change in temperature.
First, we calculate the heat gained by the water:
Q_water = m_water * c_water * ΔT_water Q_water = 50.00 g * 4.184 J/g°C * (20.15°C - 10.67°C) Q_water = 1986.96 J
Next, we calculate the heat lost by the metal. This will be equal to the heat gained by the water, but with a negative sign because the metal is losing heat:
Q_metal = -Q_water Q_metal = -1986.96 J
Finally, we rearrange the heat transfer formula to solve for the specific heat of the metal:
c_metal = Q_metal / (m_metal * ΔT_metal) c_metal = -1986.96 J / (25.00 g * (99.0°C - 20.15°C)) c_metal = 0.45 J/g°C
So, the specific heat of the metal is approximately 0.45 J/g°C.
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