If 82.4 g of cold water (initially at 22 °C) is mixed in an insulated container with 96 g of hot water (initially at 88 °C), what will be the final temperature of the water in °C?
heat gained by cold water + heat lost by hot water = 0
[mass cold H2O x specfic heat H2O x (Tfinal-Tinitial)] + [mass hot H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute the numbers and solve for Tf.
To find the final temperature of the water, we can use the principle of conservation of energy, which states that the total heat gained by one object is equal to the total heat lost by another object.
The heat gained or lost by an object can be calculated using the formula:
Q = mcΔT
Where:
Q is the heat gained or lost (in joules),
m is the mass of the object (in grams),
c is the specific heat capacity of the substance (for water, it's approximately 4.18 J/g°C),
ΔT is the change in temperature (final temperature - initial temperature).
Let's calculate the heat lost by the hot water first:
Q_hot = m_hot * c * ΔT_hot
Where:
m_hot = mass of hot water = 96 g,
ΔT_hot = temperature change of hot water = final temperature - initial temperature = T_final - 88 °C.
Now let's calculate the heat gained by the cold water:
Q_cold = m_cold * c * ΔT_cold
Where:
m_cold = mass of cold water = 82.4 g,
ΔT_cold = temperature change of cold water = final temperature - initial temperature = T_final - 22 °C.
Since the container is insulated, the heat gained by the cold water is equal to the heat lost by the hot water. So, we can set up the equation:
Q_hot = Q_cold
m_hot * c * ΔT_hot = m_cold * c * ΔT_cold
Plugging in the given values, we have:
96 g * 4.18 J/g°C * (T_final - 88 °C) = 82.4 g * 4.18 J/g°C * (T_final - 22 °C)
Now, we can solve this equation to find the value of T_final.