What will be the final temperature of the water in an insulated container as the result of passing 6.00g of steam at 100.0 degrees Celsius into 100.0g of water at 24.0 degrees Celsius?
heat lost by condensing steam + heat lost by 100 C water + heat gained by 24C water = 0
-(mass steam x heat vap) + [mass condensed steam x specific heat x (Tfinal-Tinitial)] + [mass 24 C H2O x specific heat x (Tfinal-Tinitial)] = 0
Solve for Tfinal.
To find the final temperature of the water in the insulated container, we can use the principle of conservation of energy.
First, we need to calculate the heat gained by the water:
Q_1 = m_1 * C_1 * (T_f - T_i)
Where:
Q_1 is the heat gained by the water
m_1 is the mass of water (100.0 g)
C_1 is the specific heat capacity of water (4.184 J/g°C)
T_f is the final temperature of the water (unknown)
T_i is the initial temperature of the water (24.0°C)
Next, we need to calculate the heat lost by the steam:
Q_2 = m_2 * C_2 * (T_f - T_i)
Where:
Q_2 is the heat lost by the steam
m_2 is the mass of steam (6.00 g)
C_2 is the specific heat capacity of steam (2.03 J/g°C)
T_f is the final temperature of the water (unknown)
T_i is the initial temperature of the steam (100.0°C)
Since the system is insulated, the heat gained by the water (Q_1) is equal to the heat lost by the steam (Q_2), so we can equate the two equations:
m_1 * C_1 * (T_f - T_i) = m_2 * C_2 * (T_f - T_i)
Now we can solve for T_f:
100.0 g * 4.184 J/g°C * (T_f - 24.0°C) = 6.00 g * 2.03 J/g°C * (T_f - 100.0°C)
(418.4 g°C * T_f - 418.4 g°C * 24.0°C) = (12.18 g°C * T_f - 12.18 g°C * 100.0°C)
418.4 g°C * T_f - 10046.4 g°C = 12.18 g°C * T_f - 1218.0 g°C
418.4 g°C * T_f - 12.18 g°C * T_f = 10046.4 g°C - 1218.0 g°C
(418.4 g°C - 12.18 g°C) * T_f = (10046.4 g°C - 1218.0 g°C)
406.22 g°C * T_f = 8835.4 g°C
T_f = 8835.4 g°C / 406.22 g°C
T_f ≈ 21.75°C
Therefore, the final temperature of the water in the insulated container will be approximately 21.75°C.
To find the final temperature of the water, we can use the principle of heat transfer and the equation of heat transfer. The equation for heat transfer is:
q = m * c * ΔT
Where:
q is the heat transferred
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
In this case, we have two substances: steam and water. We need to find the final temperature after they mix.
First, let's find the heat transferred from the steam to the water:
q_steam = m_steam * c_steam * ΔT_steam
Given:
m_steam = 6.00g
c_steam = 2.03 J/g°C (specific heat capacity of steam)
ΔT_steam = final temperature - initial temperature = final temperature - 100.0°C
Next, let's find the heat transferred from the water:
q_water = m_water * c_water * ΔT_water
Given:
m_water = 100.0g
c_water = 4.18 J/g°C (specific heat capacity of water)
ΔT_water = final temperature - initial temperature = final temperature - 24.0°C
Since the container is insulated, the heat transferred from steam to water will be equal to the heat transferred from water to steam:
q_steam = q_water
Using the equations above, we can set up the following equation:
m_steam * c_steam * ΔT_steam = m_water * c_water * ΔT_water
Plugging in the given values:
(6.00g) * (2.03 J/g°C) * (final temperature - 100.0°C) = (100.0g) * (4.18 J/g°C) * (final temperature - 24.0°C)
Simplifying the equation will give us the final temperature of the water.