A 44.0g sample of glass, which has a specific heat capacity of 0.670·J·g−1°C−1, is dropped into an insulated container containing 200.0g of water at 65.0°C and a constant pressure of 1atm. The initial temperature of the glass is 8.2°C.Assuming no heat is absorbed from or by the container, or the surroundings, calculate the equilibrium temperature of the water. Be sure your answer has the correct number of significant digits.
Question
A 44.0g sample of glass, which has a specific heat capacity of 0.670·J·g−1°C−1, is dropped into an insulated container containing 200.0g of water at 65.0°C and a constant pressure of 1atm. The initial temperature of the glass is 8.2°C.Assuming no heat is absorbed from or by the container, or the surroundings, calculate the equilibrium temperature of the water. Be sure your answer has the correct number of significant digits.
Solution
To solve this problem, we need to use the principle of conservation of energy, which states that energy cannot be created or destroyed, only transferred. In this case, the heat lost by the water will be equal to the heat gained by the glass.
The formula for heat transfer is Q = mcΔT, where:
- Q is the heat transferred,
- m is the mass,
- c is the specific heat capacity, and
- ΔT is the change in temperature.
For the water, the heat lost (Q_water) can be calculated as:
Q_water = m_water * c_water * (T_initial_water - T_final)
For the glass, the heat gained (Q_glass) can be calculated as:
Q_glass = m_glass * c_glass * (T_final - T_initial_glass)
Since the heat lost by the water is equal to the heat gained by the glass, we can set these two equations equal to each other and solve for the final temperature (T_final):
m_water * c_water * (T_initial_water - T_final) = m_glass * c_glass * (T_final - T_initial_glass)
Substituting the given values:
200.0g * 4.18 J/g°C * (65.0°C - T_final) = 44.0g * 0.670 J/g°C * (T_final - 8.2°C)
Solving this equation will give us the final temperature of the water.
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