Two - 20.0 g ice cubes @ -20 degrees C are placedinto275 g of water @ 25 degrees C. Assuming no energy is transfered to or from the surroundings caculate the final temperature of the water after the ice melts.
[mass ice cubes x specific heat ice x (Tfinal-Tinitial)] + [mass ice cubes x heat fusion H2O)] + [mass melted ice cubes x specific heat H2O x (Tfinal-Tinitial)] + [mass water x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for Tfinal.
To calculate the final temperature of the water after the ice melts, we need to consider the energy exchanged during the process. The energy exchanged can be calculated using the equation:
Q = mcΔT
where:
Q - heat exchanged
m - mass
c - specific heat capacity
ΔT - change in temperature
In this case, we have two ice cubes, each weighing 20.0 g and at a temperature of -20 degrees Celsius. The specific heat capacity of ice is 2.09 J/g°C.
First, let's calculate the heat exchanged when the ice cubes melt:
Q = mcΔT
Q = 40.0 g * 2.09 J/g°C * (0 - (-20)°C)
Q = 40.0 g * 2.09 J/g°C * 20°C
Q = 1672 J
The energy gained by the water through the heat exchange is equal to the energy lost by the ice cubes, so:
Qwater = -Qice
Qwater = -1672 J
Now, let's find out the final temperature of the water using the formula:
Q = mcΔT
For the water, we have mass (m) = 275 g, specific heat capacity (c) = 4.18 J/g°C (specific heat capacity of water), and we need to determine ΔT:
Qwater = mcΔT
-1672 J = 275 g * 4.18 J/g°C * (Tfinal - 25°C)
Solving for Tfinal:
Tfinal - 25°C = -1672 J / (275 g * 4.18 J/g°C)
Tfinal - 25°C = -1.9°C
Now, solving for Tfinal:
Tfinal = -1.9°C + 25°C
Tfinal = 23.1°C
Therefore, the final temperature of the water after the ice melts is 23.1 degrees Celsius.