An unknown substance has a mass of 0.125 kg and an initial temperature of 98.0°C. The substance is then dropped into a calorimeter made of aluminum containing 0.285 kg of water initially at 30.0°C. The mass of the aluminum container is 0.150 kg, and the temperature of the calorimeter increases to a final equilibrium temperature of 32.0°C. Assuming no thermal energy is transferred to the environment, calculate the specific heat of the unknown substance.
Specific heat = (0.125 kg)(98.0°C - 32.0°C) / [(0.285 kg)(32.0°C - 30.0°C) + (0.150 kg)(32.0°C - 30.0°C)]
Specific heat = 0.845 J/g°C
To calculate the specific heat of the unknown substance, we can use the equation:
Heat gained by water + heat gained by aluminum = heat lost by unknown substance
The heat gained or lost by a substance can be calculated using the formula:
Heat = mass x specific heat x change in temperature
First, let's calculate the heat gained by the water and the aluminum. The specific heat of water is 4186 J/kg°C and the specific heat of aluminum is 897 J/kg°C.
Heat gained by water = (mass of water) x (specific heat of water) x (change in temperature of water)
= (0.285 kg) x (4186 J/kg°C) x (32.0 - 30.0)°C
= 2398.76 J
Heat gained by aluminum = (mass of aluminum) x (specific heat of aluminum) x (change in temperature of aluminum)
= (0.150 kg) x (897 J/kg°C) x (32.0 - 30.0)°C
= 269.1 J
Now, let's calculate the heat lost by the unknown substance.
Heat lost by unknown substance = (mass of unknown substance) x (specific heat of unknown substance) x (change in temperature of unknown substance)
= (0.125 kg) x (specific heat of unknown substance) x (final temperature - initial temperature)
= (0.125 kg) x (specific heat of unknown substance) x (32.0 - 98.0)°C
Since no thermal energy is transferred to the environment, the heat gained by the water and the aluminum should be equal to the heat lost by the unknown substance. Therefore, we can set up the following equation:
(heat gained by water) + (heat gained by aluminum) = (heat lost by unknown substance)
2398.76 J + 269.1 J = (0.125 kg) x (specific heat of unknown substance) x (32.0 - 98.0)°C
2667.86 J = (0.125 kg) x (specific heat of unknown substance) x (-66.0)°C
We can solve this equation to find the specific heat of the unknown substance:
specific heat of unknown substance = 2667.86 J / ((0.125 kg) x (-66.0)°C)
specific heat of unknown substance = -53.32 J/kg°C (rounded to two decimal places)
Therefore, the specific heat of the unknown substance is approximately -53.32 J/kg°C.
To calculate the specific heat of the unknown substance, we can use the equation:
Q = mcΔT
Where:
Q = heat transferred
m = mass
c = specific heat
ΔT = change in temperature
First, let's calculate the heat transferred to the water and the aluminum container.
For the water:
m_water = 0.285 kg
c_water = 4.186 J/g°C (specific heat of water)
ΔT_water = (final temperature - initial temperature) = 32.0°C - 30.0°C = 2.0°C
Q_water = (m_water)(c_water)(ΔT_water)
Q_water = (0.285 kg)(4.186 J/g°C)(2.0°C)
Q_water = 2.262 J
For the aluminum container:
m_aluminum = 0.150 kg
c_aluminum = 0.897 J/g°C (specific heat of aluminum)
ΔT_aluminum = (final temperature - initial temperature) = 32.0°C - 30.0°C = 2.0°C
Q_aluminum = (m_aluminum)(c_aluminum)(ΔT_aluminum)
Q_aluminum = (0.150 kg)(0.897 J/g°C)(2.0°C)
Q_aluminum = 0.269 J
Now, let's calculate the heat transferred to the unknown substance.
Q_substance = Q_water + Q_aluminum
Q_substance = 2.262 J + 0.269 J
Q_substance = 2.531 J
Finally, let's calculate the specific heat of the unknown substance.
Q_substance = (m_substance)(c_substance)(ΔT_substance)
2.531 J = (0.125 kg)(c_substance)(32.0°C - 98.0°C)
2.531 J = (0.125 kg)(c_substance)(-66.0°C)
Dividing both sides by (0.125 kg)(-66.0°C), we get:
c_substance = 2.531 J / [(0.125 kg)(-66.0°C)]
c_substance = 2.531 J / (-0.00825 kg·°C)
Finally, we can calculate the specific heat of the unknown substance:
c_substance ≈ -306.85 J/kg·°C
Note: The negative sign indicates that the unknown substance has absorbed heat from the water and the aluminum container.