A 59.0g sample of polystyrene is put into a calorimeter (see sketch at right) that contains 300.0g of water. The polystyrene sample starts off at 86.5°C and the temperature of the water starts off at 20.0°C. When the temperature of the water stops changing it's 24.1°C. The pressure remains constant at 1atm.Calculate the specific heat capacity of polystyrene according to this experiment. Be sure your answer is rounded to the correct number of significant digits.
Question
A 59.0g sample of polystyrene is put into a calorimeter (see sketch at right) that contains 300.0g of water. The polystyrene sample starts off at 86.5°C and the temperature of the water starts off at 20.0°C. When the temperature of the water stops changing it's 24.1°C. The pressure remains constant at 1atm.Calculate the specific heat capacity of polystyrene according to this experiment. Be sure your answer is rounded to the correct number of significant digits.
Solution
To calculate the specific heat capacity of polystyrene, we need to use the formula for heat transfer:
q = mcΔT
where: q = heat energy m = mass c = specific heat capacity ΔT = change in temperature
First, we need to calculate the heat gained by the water, which is the same as the heat lost by the polystyrene (since the system is isolated).
The mass of the water (m_water) is 300.0g, the specific heat capacity of water (c_water) is 4.184 J/g°C, and the change in temperature of the water (ΔT_water) is 24.1°C - 20.0°C = 4.1°C.
So, the heat gained by the water (q_water) is:
q_water = m_water * c_water * ΔT_water q_water = 300.0g * 4.184 J/g°C * 4.1°C q_water = 5143.44 J
This is the heat lost by the polystyrene, so q_polystyrene = -5143.44 J (the negative sign indicates heat loss).
The mass of the polystyrene (m_polystyrene) is 59.0g, and the change in temperature of the polystyrene (ΔT_polystyrene) is 86.5°C - 24.1°C = 62.4°C.
We can now calculate the specific heat capacity of the polystyrene (c_polystyrene) using the formula:
c_polystyrene = q_polystyrene / (m_polystyrene * ΔT_polystyrene) c_polystyrene = -5143.44 J / (59.0g * 62.4°C) c_polystyrene = -1.39 J/g°C
So, the specific heat capacity of polystyrene according to this experiment is -1.39 J/g°C. The negative sign indicates that the polystyrene lost heat. However, specific heat capacity is a scalar quantity and should always be positive. Therefore, the absolute value is taken, giving a specific heat capacity of 1.39 J/g°C.
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