A student following the procedure in this experiment acquired temperature as a f

Question

A student following the procedure in this experiment acquired temperature as a function of time data in order to determine the heat capacity of the calorimeter. Based on a plot of these data and extrapolation back to time zero, the student determined the final temperature to the 14.3 degree C. The room temperature water had a volume of 49.2 mL. The cold water had a volume of 49.7 mL. The student measured T_h = 22.4 degree C and T_c= 2.9 degree C. Use these data to determine the heat capacity of the calorimeter, in J/degree C.

Explanation / Answer

Tfinal = 14.3 °C

V = 49.2 mL

V = 49.7 mL

Th = 22.4°C, Tc = 2.9°C

find Capacity of calorimeter

Qcold + Qhot + Qcalorimter = 0

Qcold = mcold * Cwater * (Tf-Tcold)

Qhot = mhot * Cwater * (Tf-Thor)

Qcal= Ccal * (Tf-Troom)

mcold * Cwater * (Tf-Tcold) + mhot * Cwater * (Tf-Thor) + Ccal * (Tf-Troom) = 0

49.2 * 4.184* (14.3-2.9) + 49.7* 4.184* (14.3-22.4) + Ccal * (14.3-25) = 0

Solve for Ccal

2346.721 + -1684.35 + Ccal *-10.7= 0

Ccal = (2346.721 + -1684.35) / (10.7) = 61.90 J/°C

Ccal = 61.90 J/°C

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