Question

A 16.260g piece of metal was heated in a hot water bath at
95.5°C
. The hot metal was then transferred to a calorimeter containing 50.00 mL of water (d of H2O= 1.00 g/mL). From the time-temperature plot, the initial and final temperature for the water were determined to be and , respectively.

A) calculate heat (q) gained by the calorimeter and water assuming Ccal= , qH2O= (4.184 J/gC)(mwater)() and qcal=Ccal.

Answer 1

Mass of water, m= 50.00 mLx3 = 50.00 g Specificheat of water ,s = 4.184 J/g. C Change in temperature A7 = 26.18°C-23.76°C = 2.42°C Amount of heat absorbedby water, water = mSAT = 50.00 8x4,184X(242°C) = 506.3J Change in temperature produced, A7 = 2.42°C Heat capacity of calorimeter ,s = 21.0 J/°C Amount of heat absorbed by the calorimeter, cal = CAT =(210 )(242°c) = 50.82 J