5.25g of unknown metal at 990C was placed into 15.0 g of water at 250C. The mixture came to a final temperature of 410C. What is the specific heat of the metal if the specific heat of water is 4.184 J/g0C?
Question
5.25g of unknown metal at 990C was placed into 15.0 g of water at 250C. The mixture came to a final temperature of 410C. What is the specific heat of the metal if the specific heat of water is 4.184 J/g0C?
Solution
To solve this problem, we can use the principle of conservation of energy, which states that the heat lost by the hot object (the metal) is 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 capacity 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 = 15.0g * 4.184 J/g°C * (41°C - 25°C) Q_water = 1005.36 J
The heat lost by the metal is equal to the heat gained by the water, so:
Q_metal = Q_water Q_metal = 1005.36 J
Then, we can rearrange the formula Q = mcΔT to solve for the specific heat capacity of the metal (c_metal):
c_metal = Q_metal / (m_metal * ΔT_metal) c_metal = 1005.36 J / (5.25g * (99°C - 41°C)) c_metal = 1005.36 J / (5.25g * 58°C) c_metal = 0.31 J/g°C
So, the specific heat of the metal is approximately 0.31 J/g°C.
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