A copper calorimeter can with mass 0.202 kg contains 0.185 kg of water and 1.40×10^−2 kg of ice in thermal equilibrium at atmospheric pressure. If 0.900 kg of lead at a temperature of 417 degrees celsius is dropped into the can, what is the final temperature of the system if no heat is lost to the surroundings?
To find the final temperature of the system, we can use the principle of conservation of energy, assuming that no heat is lost to the surroundings.
The heat gained by the ice can be calculated using the formula:
Q_ice = m_ice * c_ice * ΔT_ice
Where:
m_ice is the mass of the ice (0.014 kg)
c_ice is the specific heat capacity of ice (2.09 kJ/kg·K)
ΔT_ice is the change in temperature of the ice
Since the ice is in thermal equilibrium, the change in temperature is zero, so the heat gained by the ice is also zero.
The heat gained by the water can be calculated using the formula:
Q_water = m_water * c_water * ΔT_water
Where:
m_water is the mass of the water (0.185 kg)
c_water is the specific heat capacity of water (4.18 kJ/kg·K)
ΔT_water is the change in temperature of the water
Again, since the water is in thermal equilibrium, the change in temperature is zero, so the heat gained by the water is also zero.
Now, let's consider the heat gained by the lead:
Q_lead = m_lead * c_lead * ΔT_lead
Where:
m_lead is the mass of the lead (0.900 kg)
c_lead is the specific heat capacity of lead (0.13 kJ/kg·K)
ΔT_lead is the change in temperature of the lead
Since the lead starts at a temperature of 417 degrees Celsius, we need to convert it to Kelvin:
T_lead_initial = 417 + 273.15 = 690.15 K
Assuming the final temperature of the system is T_final, we can set up an equation:
Q_lead = Q_ice + Q_water
m_lead * c_lead * (T_final - T_lead_initial) = 0 + 0
0.900 * 0.13 * (T_final - 690.15) = 0
Simplifying the equation:
0.117 * (T_final - 690.15) = 0
T_final - 690.15 = 0
T_final = 690.15
Hence, the final temperature of the system, if no heat is lost to the surroundings, is 690.15 Kelvin.