Question

1.Use the recorded data to fill in the blanks. Show work for calculations.

		Trial 1	Trial 2	Trial 3
a	Initial temperature of the cold water in the calorimeter (°C):	21.5	21.5	21.5
b	Initial temperature of the hot water (°C):	60.0	60.0	60.0
c	Mass of the hot water (assume density of 1.00 g/mL) (g)	50.0g	50.0g	50.0g
d	Mass of the cold water (assume density of 1.00 g/mL) (g)	50.0g	50.0g	50.0g
e	Final (maximum) temperature of water mixture the calorimeter (°C):	39.8	39.8	39.8
f	Calculate change in Temperature, DThot water (Tfinal – Tinitial hot)	-10.2	-10.2	-10.2
g	Calculate change in temperature, DTcold water (Tfinal – Tinitial cold)	18.3	18.3	18.3
h	Calculate qhot (J)
j	Calculate qcold (J)
k	Calculate qcalorimeter (J)
l	Calculate Heat Capacity of Calorimeter, Cp(J/°C)
	Average Heat Capacity of Calorimeter, Cp, J/°C)

In another industrial lab, the heat capacity of the calorimeter was measured by a more precise method, and was determined to be 21.3 J/°C. Note because C_p values are so small, your experimental value may deviate largely (up to about 30% error) from the industrial value determined with more precise equipment. Use this industrial value (theoretical value) and the average C_p value you obtained in Experiment 1 (experimental value) to determine the percent error in your experimental value.:

Answer 1

Calculate change in Temperature, DT_{hot water} = (T_final – T_{initial hot}) = 39.8 - 60 = -20.2

q_hot = m x c x (DT)_{hot water}

q_hot = 50 x 4.18 x (-20.2)

q_hot = -4221.8 J

q_cold = m x c x (DT)_{cold water}

q_cold = 50 x 4.18 x (18.3)

q_cold = 3824.7 J

Heat lost by hot water = heat gained by calorimeter + heat gained by cold water

q_hot = qcalorimeter + q_cold

q_calorimeter = q_hot - q_cold

q_calorimeter =4221.8 -3824.7 = 397.1 J

q_calorimeter = 397.1 J

Calorimeter contains cold water

so specific heat of calorimeter =

c=397.1/18.3 = 21.7 J/^oC

specific heat of calorimeter = 21.7 J/^oC

percentage error =

percentage error =

percentage error = 1.9%