Ice cubes of total mass 6.00 grams have temperature -4.0°C, and are placed in 50.0 grams of water at 20.0°C in an insulated foam cup. the equilibrium temperature of the water inside the cup? READ DETAILS, TOO.
Assume that the system is isolated For water Lf= 334kJ/kg and Lv= 2257kJ/kg The specific heat capacity of ice is 2.05kJ/kg * Celcius, and for liquid water is 4.19kJ/kg * C .
9.1°C
11.4°C
14.8°C
5.5°C
these are the answer choices.
Assume that the system is isolated For water Lf= 334kJ/kg and Lv= 2257kJ/kg The specific heat capacity of ice is 2.05kJ/kg * Celcius, and for liquid water is 4.19kJ/kg * C .
9.1°C
11.4°C
14.8°C
5.5°C
these are the answer choices.
1 Answer
This is what I get
Explanation:
Let equilibrium temperature be
A. Heat gained by ice to become water at
- Heat gained by ice at
#-4.0^@\ C# to become ice at#0^@\ C# #m_is_iDeltaT=6.00/1000xx2.05xx4=0.0492\ kJ# - Heat gained by ice at
#0^@\ C# to become water at#0^@\ C# #m_iL_f=6.00/1000xx334=2.004\ kJ# - Heat gained by water at
#0^@\ C# to become water at#t^@\ C# #m_is_wDeltaT=6.00/1000xx4.19xxt=0.02514t\ kJ#
Total gained by ice#Q_g=(2.0532+0.02514t)\ kJ#
B. Heat lost by water at
#Q_l=m_ws_wDeltaT=50.0/1000xx4.19xx(20-t)=0.2095(20-t)#
#=>Q_l=(4.19-0.2095t)\ kJ#
From Law of conservation of Energy for a closed system.
#Q_l=Q_g#
#:.4.19-0.2095t=2.0532+0.02514t#
#=>0.2095t+0.02514t=-2.0532+4.19#
#=>0.23464t=2.1368#
#=>t=9.1^@\ C#