Can you please type the explanation to the questions

\fTable of data TRIAL 1 TRIAL 2 TRIAL 3 Mass of calorimeter cup, mc 60.8g 60.8g 60.8g Volume of tap water V , mL or cm- 140 mL 170 mL 190 mL Mass of tap water my. kg 0.14 kg 0.17 kg 0.19 kg Initial Temperature of cup and water Ti, C 30 RC 25 RC 26 RC Final Temperature of cup, water and melted ice If, 15 RC 19 RC 18 AC OC Volume of ice added Vi, mL 19 mL 21 mL 18 mL Mass of ice added mi, kg 0.019 kg 0.021 kg 0.018 kg Lf kcal/kg 106.086 kcal/kg 33.39 kcal/kg 72.39 kcal/kgANALYSIS: We will use the relation: Qcold - - Qhot as we did in class. Q cold is the heat gained to melt the ice which is initially at 0 C and warm it to a final equilibrium temperature, Tf. Qhot is the amount of heat given off by the tap water and calorimeter cup from an initial temperature of Ti to a final equilibrium temperature of Tf. REMEMBER that the change in temperature is final temperature - initial temperature. This becomes: Qcold = - Qhot milf + micw(Tf - 0) = -[mwcw(Tf - Ti) + mccc(Tf - Ti)] Equation (1) where cw is the specific heat of water = 1.0 kcal/kg .C, and cc=0.22 kcal/kg .C is the specific heat of the aluminum calorimeter cup. We simplify this equation to become milf + micwTf = mwcw(Ti- Tf) + mccc(Ti - If) This can be rewritten as milf = MWCw(Tf - Ti) + mccc(Ti - Tf) - micwIf Le = mucw (T; - Ty) + mcc (T; - T,) - m,c,If Equation (2) m; OR Le = [ww + mcc ](T; - T; ) - m,c,If Equation (3) m; We can thus determine If using the values of our measurements for each trial. We should find the average value for the three trials and compare with the known value of 80 kcal/kg by calculating the percentage error. (Remember: Cw = 1.0 kcal/kg .C, and cc = 0.22 kcal/kg .C)