The final temperature of the mixture is 46.2°C.
According to the theory of heat exchange, heat absorbed by ice is equal to the heat given by water. Ice absorbs heat to first melt at a constant temperature of 0°C and then water at 0°C increases its temperature to the final temperature T. The amount of heat for this process is provided by hot water at 50°C, which cools down to the final temperature of the mixture T.
Heat absorbed by ice is given by,[tex] [tex] Q_{a} =m_{ice} *L+m_{ice}*c_{w} (T-0) [/tex]
=(1 kg)(3.34*10^5 J/K)+(1 kg)(4160 J/kg)(T-0)
=3.34*10^5+4160T
[/tex]
heat given by water is given by
[tex] Q=m_{w} *c_{w} *(T_{w} -T)
=(9kg)(4160 J/kg)(50-T) [/tex]
Equating both the equations and solving for T,
°[tex] 33280 T=1.538*10^6
T=46.2 C [/tex]
ice:
mass (m) = 1 kg
Initial Temperature (T) = 0 ° C
Latent heat (L) = 3.34 * 105 J/kg
Water:
mass (m) = 9 kg
Initial Temperature (T) = 50 ° C
Specific heat capacity (c) = 4,160 J/kg
First you must find the amount of heat required for the ice to turn into water (Q is heat)
Q = mL
Q = 1 * ( 3.34 * [tex]10^{5}[/tex])
Q = 3.34 * [tex]10^{5}[/tex] J
1 kg of ice needs 3.34 * J of energy to convert into water
The heat lost by the water is equal to the heat gained by the ice.
Heat can be found using the equation: Q = mcΔT therefore...
mcΔT (water) = mcΔT (ice)
9 * 4,160(50 - x) = 1 * 4,160(x - 0) + 3.34 * [tex]10^{5}[/tex]
You need to add the heat needed for the ice to convert into the water (3.34 * [tex]10^{5}[/tex] J) to the ice side of the equation, because when we get the final temperature the ice will be converted into water.
x = 36.97
The resulting temperature is 36.97 ° C
I hope this helped!
~Just a girl in love with Shawn Mendes
