A 2kg mass of lead at 100°C is dropped into a copper vessel containing 0.3kg of water at 0°C and rapidly stirred.The final temperature attained by the vessel and its content is 16°C .Take specific heat capacity of lead as 130J/kgk, find the heat capacity of the copper vessel,
To find the heat capacity of the copper vessel, we need to calculate the amount of heat transferred to both the lead and the water and then subtract that from the total heat added to find the heat capacity of the vessel.
First, let's calculate the heat transferred to the lead:
Q_lead = m_lead * c_lead * ΔT_lead
Q_lead = 2kg * 130J/kgK * (16°C - 100°C)
Q_lead = 2kg * 130J/kgK * (-84°C)
Q_lead = -21,120J
Next, let's calculate the heat transferred to the water:
Q_water = m_water * c_water * ΔT_water
Q_water = 0.3kg * 4186J/kgK * (16°C - 0°C)
Q_water = 0.3kg * 4186J/kgK * 16°C
Q_water = 20,104.8J
The total heat added to the system is:
Q_total = Q_lead + Q_water
Q_total = -21,120J + 20,104.8J
Q_total = -1,015.2J
Now, we can calculate the heat capacity of the copper vessel:
C_vessel = Q_total / ΔT_total
C_vessel = -1,015.2J / (16°C - 0°C)
C_vessel = -1,015.2J / 16°C
C_vessel = -63.45J/°C
The negative sign indicates that the copper vessel lost heat during the process. The heat capacity of the copper vessel is approximately 63.45J/°C.