A large punch bowl holds 3.35 kg of lemonade (which is essentially water) at 20.0° C. A 0.28kg ice cube at
−10.2°
C is placed in the lemonade. What is the final temperature of the system, and the amount of ice (if any) remaining? Ignore any heat exchange with the bowl or the surroundings.
__° C
__ kg
i tried doing Q_w = (m_water)(c_water) (delta T)
then Q_ice = m_ice L_f
then getting the difference of those and dividing it over (mwater+mice)(specific heat of water)(delta T)
but it isnt working. help?
−10.2°
C is placed in the lemonade. What is the final temperature of the system, and the amount of ice (if any) remaining? Ignore any heat exchange with the bowl or the surroundings.
__° C
__ kg
i tried doing Q_w = (m_water)(c_water) (delta T)
then Q_ice = m_ice L_f
then getting the difference of those and dividing it over (mwater+mice)(specific heat of water)(delta T)
but it isnt working. help?

specific heat of water is 4.186 kJ/kgC
specific heat of ice is 2.06 kJ/kgC
heat of fusion of ice is 334 kJ/kg
Energy given up by the ice in melting and warming up to the final temp, T, equals that of the lemonade cooling off from 29º to T, assuming all of the ice has melted.
E1 = 2.06×0.28×10.2 + 334×0.28 + 4.186×0.28×T
E2 = 4.186×3.35×(20–T)
E1 = 5.88 + 93.52 + 1.17T = 99.4 + 1.17T
E2 = 280.46 – 14.02T
99.4 + 1.17T = 280.46 – 14.02T
15.19T = 181.1
T = 11.9º
.
specific heat of ice is 2.06 kJ/kgC
heat of fusion of ice is 334 kJ/kg
Energy given up by the ice in melting and warming up to the final temp, T, equals that of the lemonade cooling off from 29º to T, assuming all of the ice has melted.
E1 = 2.06×0.28×10.2 + 334×0.28 + 4.186×0.28×T
E2 = 4.186×3.35×(20–T)
E1 = 5.88 + 93.52 + 1.17T = 99.4 + 1.17T
E2 = 280.46 – 14.02T
99.4 + 1.17T = 280.46 – 14.02T
15.19T = 181.1
T = 11.9º
.