Change in Initial Energy

Question

Calculate the mass of ice (c= 2.11 J/gC) required to melt (initial temperature 0 Celsius) in a 500 gram sample of iced team (assume c= 4.184 J/g C since it is mostly water) to lower the temperature of the tea from 25 Celsius to 20 Celsius. Assume that once the ice melts it reaches a final temperature of 20 Celsius as well. Also assume that none of the heat is lost to the surroundings but is locked inside the Styrofoam cup and lid.

I'm uncertain as to how to do the math for these equations, and it would be great if someone could help me understand them better!

Answer 1

heat gained by the ice = heat lost by the tea

m_iceC_ice∆T_ice = m_teaC_tea∆T_tea

(m_ice)(2.11 J/gC)(20 - 0 C) = (500 g)(4.184 J/gC)(25 - 20 C)

(m_ice)(2.11 J/gC)(20 C) = (2092 J/C)(5 C)

(m_ice)(42.2 J/g) = 10,460 J

m_ice = 10,460 J/42.2 J/g = 248 g

(NOTE: Many text books and teachers use heat gained by ice = -heat lost by tea, and then use ∆T as T_final-T_initial. I prefer to avoid the negative signs by doing it the way I did it above and always keeping the ∆T as a positive value).