if 75 g of ice at 0.0 degrees C were added to 1.5 L of water at 75 degrees C what would be the final temperature of the mixture
To melt ice requires q1 = mass ice x heat fusion.
To raise T of melted ice at zero C to Tfinal requires q2 = mass melted ice x specific heat liquid H2O x (Tfinal-Tinitial)
To cool the water from 75C to Tfinal requires q3 = mass liquid water x specific heat x (Tfinal-Tinitial).
Just substitute q1 + q2 + q3 = 0 and solve for Tf.
how many molecules of h2o if formed from this equation Al2O3 + 3H2 ->2Al + 3H2O
Beats me
To calculate the final temperature of the mixture, we can use the principle of conservation of energy.
First, let's determine the heat gained or lost by each component of the system:
1. Heat gained or lost by ice (Qice):
Qice = (mass of ice) x (specific heat capacity of ice) x (change in temperature)
= (75 g) x (2.09 J/g·°C) x (final temperature - 0.0°C)
2. Heat gained or lost by water (Qwater):
Qwater = (mass of water) x (specific heat capacity of water) x (change in temperature)
= (1.5 kg) x (4.18 J/g·°C) x (final temperature - 75°C)
According to the principle of conservation of energy, the heat lost by the warmer substance (water) is equal to the heat gained by the colder substance (ice):
Qwater = -Qice
Now we can set up the equation:
(1.5 kg) x (4.18 J/g·°C) x (final temperature - 75°C) = -(75 g) x (2.09 J/g·°C) x (final temperature - 0.0°C)
Simplifying the equation:
(1.5 kg) x (4.18 J/g·°C) x final temperature - (1.5 kg) x (4.18 J/g·°C) x 75°C = -(75 g) x (2.09 J/g·°C) x final temperature
Solving for the final temperature:
(6.27 kg·°C) x final temperature - (6.27 kg·°C) x 75°C = -(0.07875 kg·°C) x final temperature
6.27 kg·°C x final temperature - 6.27 kg·°C x 75°C = -0.07875 kg·°C x final temperature
(6.27 kg·°C + 0.07875 kg·°C) x final temperature = 6.27 kg·°C x 75°C
6.34875 kg·°C x final temperature = 470.25 kg·°C
final temperature = 470.25 kg·°C / 6.34875 kg·°C
final temperature ≈ 74.1°C
Therefore, the final temperature of the mixture would be approximately 74.1°C.