If 7.20 kg of ice at -5.00°C is added to 12.0 kg of water at 30°C, compute the final temperature.
The sum of heats added is zero .
heat added to ice+heatadded to water=0
first assumption: assume all the ice melts, if there is a solution below zero, you know that can't be .
heating up ice+heat ice melting+heat ice water heating+heatoriginal water=0
7.20*Cice*(0-(-5)+ 7.30*Hf+7.30Cwater(Tf-0)+12*cwater*(Tf-30)=0
Now do the math, and solve for Tf. Look up the specific heats for ice and water, and the heat of fusion for water.
To compute the final temperature when ice is added to water, we can use the principle of conservation of energy, which states that the total energy of a closed system remains constant.
First, we need to calculate the energy gained or lost by each substance during the process.
For the ice to reach its melting point at 0°C, it needs to absorb heat energy to raise its temperature from -5.00°C to 0°C. The energy gained by the ice can be calculated using the specific heat capacity of ice (C_ice), which is 2.09 J/g°C, and its mass (m_ice), which is 7.20 kg.
Q_ice = m_ice * C_ice * (final temperature - initial temperature)
Since we want to find the final temperature, we rearrange the equation:
final temperature = (Q_ice / (m_ice * C_ice)) + initial temperature
Now let's calculate the energy gained by the ice:
Q_ice = (7.20 kg) * (2.09 J/g°C) * (0°C - (-5.00°C))
Next, we need to calculate the energy gained or lost by the water. The energy lost by the water is equal to the energy gained by the ice because the total energy within a closed system remains constant.
Q_water = -Q_ice
Now we can calculate the energy gained by the water and find its final temperature:
Q_water = (m_water * C_water * (final temperature - initial temperature))
Since we know the initial temperature of the water is 30°C and the final temperature is unknown, we rearrange the equation to solve for the final temperature:
final temperature = (Q_water / (m_water * C_water)) + initial temperature
Given that the mass of water (m_water) is 12.0 kg and the specific heat capacity of water (C_water) is 4.18 J/g°C, we can substitute these values into the equation.
Finally, substitute the calculated values into the equation and solve for the final temperature.