- #1
toothpaste666
- 516
- 20
Homework Statement
An aluminum cup (of negligible mass) is filled with 0.5 kg of water with a temperature of 15◦C and 1.3 kg of ice (at -8 celsius) is added.
A)What is the final phase or phases of the mixture?
B)What is the final temperature of the mixture?
C)How much heat is required to raise the mixture to 15◦C?
Homework Equations
Q = mcdeltaT
Q = mL
The Attempt at a Solution
I guess my main confusion would be how to predict the phase it would end up in, how to tell if you were wrong , and if you are wrong what to do next. This is my solution for part A and B but I think my result is wrong.
I assumed the ice melts to water and is brought to a final temp
[itex] heat gained by ice = heat lost by water [/itex]
[itex] Q_{ice to 0 C} + Q_{ice to water} + Q_{water to T_f} = Q_{water to T_f} [/itex]
[itex] m_i c_i(0-(-8)) + m_i L_F + m_i c_w(T_f - 0) = m_w c_w(15-T_f) [/itex]
[itex] m_i[c_i(8) + L_F + c_w(T_f)] = m_w c_w (15 - T_f) [/itex]
mass of ice mi = 1.3kg, spec heat of ice ci = 2100 J/kgC, mass water mw = .5 kg, cw = 4186 J/kgC, heat of fusion of water = 33300 J/kg
[itex] 1.3[2100(8) + 33300 + 4186T_f] = (.5)(4186)(15-T_f) [/itex]
[itex] 1.3(50100 + 4186T_f) = 2093(15 - T_f) [/itex]
[itex] 65130 + 5442T_f = 31395 - 2093T_f [/itex]
[itex] 7535T_f = -33735 [/itex]
[itex] T_f = -4.48 C [/itex]
but if this was the case the ice would not have even started melting yet, it would just have lowering in temperature a few degrees. also I think it means all the water would have frozen? I am not sure how to fix this result