Calculate the final temperature of a mixture of 200.0g of ice initially at -16.5 Celsius and 406.0g of water initially at 84.5 degrees Celsius.
q1 = heat to raise temperature of ice from -16.5 C to zero C.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial)
q2 = heat to melt the ice.
q2 = mass melted ice x heat fusion
q3 = heat to raise temperature of melted ice from zero C to final T.
q3 = mass melted ice x specific heat water x (Tfinal-Tinitial)
q4 = heat released by warm water at 84.5 C when lowered to final T.
q4 = mass warm water x specific heat H2O x (Tfinal-Tinitial)
Substitute for each of the above and set to zero as
q1 + q2 + q3 = 0
and solve for Tfinal.
Post your work if you get stuck.
thank you
what about q4
To find the final temperature of a mixture, we need to use the principle of conservation of energy, known as the heat transfer equation. The equation can be expressed as:
Q = mcΔT
Where:
Q = Heat transferred (in joules)
m = Mass (in grams)
c = Specific heat capacity (in joules per gram per degree Celsius)
ΔT = Change in temperature (in degrees Celsius)
First, let's find the heat transferred when the ice melts:
Q_ice = m_ice * c_ice * ΔT_ice
Where:
m_ice = Mass of ice (200.0g)
c_ice = Specific heat capacity of ice (2.09 J/g°C)
ΔT_ice = Final temperature of ice - Initial temperature of ice = 0°C - (-16.5°C) = 16.5°C
Q_ice = 200.0g * 2.09 J/g°C * 16.5°C
Q_ice = 6639 J (joules)
Next, let's find the heat transferred when the water cools down:
Q_water = m_water * c_water * ΔT_water
Where:
m_water = Mass of water (406.0g)
c_water = Specific heat capacity of water (4.18 J/g°C)
ΔT_water = Final temperature of water - Initial temperature of water = 0°C - 84.5°C = -84.5°C
Q_water = 406.0g * 4.18 J/g°C * (-84.5°C)
Q_water = -143082 J (joules) (Note: Negative sign indicates heat loss)
Now, we need to assume that the heat lost by the water (Q_water) is equal to the heat gained by the ice (Q_ice) during the mixing process. Therefore, we can write:
Q_water = -Q_ice
-143082 J = 6639 J
Now, let's find the final temperature, which is the same for both substances:
Q = mcΔT
For the water:
-143082 J = 406.0g * 4.18 J/g°C * ΔT
ΔT = -143082 J / (406.0g * 4.18 J/g°C)
ΔT ≈ -86.4°C
For the ice:
6639 J = 200.0g * 2.09 J/g°C * ΔT
ΔT = 6639 J / (200.0g * 2.09 J/g°C)
ΔT ≈ 15.9°C
Since the final temperature is the same for both substances, it can be either -86.4°C or 15.9°C. However, since the final temperature should be physically meaningful, we can conclude that the final temperature of the mixture is approximately 15.9°C.