if 5.25g of ice AT 0 c IS PLACED IN 25.63 G OF LIQUID WATER AT 28c, WHAT WILL BE THE FINAL TEMPERATURE OF THIS MIXTURE?
cool water + melt ice + heat water from ice = 0
[mass water x specific heat water x (Tfinal-Tinitial)] + [mass ice x heat fusion] + [mass ice from water x specific heat water x (Tfinal-Tinitital)] = 0
Solve for Tfinal.
Check my work.
what would the mass be for ice from water and the specific heat water? That is what I cant seem to figure out.
Thank you
nevermind, I got it!
Thanks
hii , how are you suppose to find the mass of ice from water ?
To find the final temperature of the mixture, we can apply the principle of energy conservation.
Step 1: Calculate the heat gained by the ice as it warms up to the final temperature.
We can use the equation Q = mcΔT, where Q represents heat energy, m represents mass, c represents specific heat capacity, and ΔT represents the change in temperature.
The specific heat capacity of ice is approximately 2.09 J/g°C.
Q_ice = m_ice * c_ice * ΔT
Given:
m_ice = 5.25 g
c_ice = 2.09 J/g°C
ΔT = Final temperature - initial temperature = Final temperature - 0°C
Step 2: Calculate the heat lost by the water as it cools down to the final temperature.
Similarly, we can use the same equation mentioned earlier.
The specific heat capacity of water is approximately 4.18 J/g°C.
Q_water = m_water * c_water * ΔT
Given:
m_water = 25.63 g
c_water = 4.18 J/g°C
ΔT = 28°C - Final temperature
Step 3: Set the heat gained by the ice equal to the heat lost by the water.
Q_ice = Q_water
m_ice * c_ice * ΔT = m_water * c_water * ΔT
Cancelling out ΔT from both sides of the equation, we get:
m_ice * c_ice = m_water * c_water
Step 4: Now we can substitute the given values into the equation and solve for the final temperature.
5.25g * 2.09 J/g°C = 25.63g * 4.18 J/g°C
10.9425 J/°C = 106.9134 J/°C
Dividing both sides by 10.9425, we get:
1 = 106.9134 J/°C / 10.9425 J/°C
1 ≈ 9.77
Therefore, the final temperature is approximately 9.77°C.