化學的問提 (還滿難的)

Question

A student collected the following data for the calorimeter constant experiment :
Volume of cold water: 4.73 mL
Temperature of cold water: 9.1 °C
Volume of hot water: 7.44 mL
Temperature of hot water: 50.1 °C
Final temperature after mixing: 26.7 °C

Use this data to determine the calorimeter constant (Ccal) to three significant figures.

Assume the specific heat is 4.184 J g-1 °C-1 for both cold and hot water. Also assume the density of water is 1.00 g/mL for the hot and cold water.

Answer 1

1.for cold water：a.Volume of cold water: 4.73 mL  = 4.73 mL*1.00 g/mL = 4.73 gb.Temperature of cold water: 9.1 °Cc.H = mSΔT = (4.73)*1*(26.7-9.1) = 83.248(cal) = 348.31(j)2. for hot watera.Volume of hot water: 7.44 mL = 7.44 mL*1.00 g/mL = 7.44 gb.Temperature of hot water: 50.1 °Cc.H = mSΔT = (7.44)*1*(50.1-26.7) = 174.096(cal) = 728.42(j)3.所散失的熱量 = 728.42(j) - 348.31(j) = 380.11(j)散失熱量所佔的比率：(380.11/728.42)*100% = 52.18%