If 50 g of ice at −5
◦C is mixed into 300 g of water at 60◦C in a completely insulated container,
what is the final equilibrium temperature of the water? Why did we insist on the container
being completely insulated?
For this kind of set up, we have summation of all energy transferred equal to zero. Note that for ice, Q is different from each process: (1) from T=-5 C to T=0 ; (2) phase change of ice to water at constant T ; (3) from T=0 to final T.
For Ice:
(1) from T=-5 C to T=0. We use the formula:
Q = mcΔT
where
m = mass (g)
c = specific heat capacity (J/g-K)
T = temperature (K)
Q = ( 50 g )( 2.01 J/g-K )( 0 - (-5) K)
Q = 502.5 J
(2) phase change of ice to water at constant T
Q = mΔHf
where Hf is the latent heat of fusion
Q = ( 50 g )( 333.33 J/g )
Q = 16666.67 J
(3) from T=0 to final T. At this point, ice is now water:
Q = mcΔT
Q = ( 50 g )( 4.184 J/g-K )(T - 0)
Q = 209.2T
For hot water initially at 60 C:
Q = mcΔT
Q = ( 300 g )( 4.184 J/g-K )(T - 60)
Q = 1255.2(T - 60)
Combining all Q's:
502.5 + 16666.67 + 209.2T + 1255.2(T - 60) = 0
Now solve for T. Units in degree Celsius.
The system must be insulated so that no energy could escape into the surroundings, therefore heat absorbed and heat lost are only happening with ice and hot water.
By the way the values for latent heat and heat capacities can be found on google or textbooks.
hope this helps~ `u`
To find the final equilibrium temperature of the water, we can use the principle of conservation of energy. We know that energy gained by the ice is equal to the energy lost by the water, assuming no energy is lost to the surroundings due to the complete insulation of the container.
First, let's calculate the energy gained by the ice using the equation:
Q_ice = m_ice * c_ice * ΔT_ice
Where:
Q_ice is the energy gained by the ice,
m_ice is the mass of the ice (50 g),
c_ice is the specific heat capacity of ice (2.09 J/g°C), and
ΔT_ice is the temperature change of the ice (final equilibrium temperature - initial temperature).
Now, let's calculate the energy lost by the water using the equation:
Q_water = m_water * c_water * ΔT_water
Where:
Q_water is the energy lost by the water,
m_water is the mass of the water (300 g),
c_water is the specific heat capacity of water (4.18 J/g°C), and
ΔT_water is the temperature change of the water (final equilibrium temperature - initial temperature).
Since energy is conserved, we can equate Q_ice to Q_water:
m_ice * c_ice * ΔT_ice = m_water * c_water * ΔT_water
Now, let's solve for the final equilibrium temperature (ΔT_water):
ΔT_water = (m_ice * c_ice * ΔT_ice) / (m_water * c_water)
The reason we insisted on the container being completely insulated is to ensure that no heat is exchanged with the surroundings. If there was any heat exchange, it would affect the final equilibrium temperature and make our calculations inaccurate. By having a completely insulated container, we can assume that the system is isolated and the only heat exchange is between the ice and water, leading to an accurate calculation of the final equilibrium temperature.