Question

please help me calculate

Data Tria Trial (a) Mass of empty Styrofoam cups 10.55 10.63 79 15 79. 70 (b) Mass of cups +70 mL water 102.51 (c) Mass of cups + 70 mL water+30 mL hot 107. P 23. (d) Initial temperature of water in the calorimeter 3 °C °C 92.0 °C (e) Temperature of the boiling water bath °C 430 43.5 (f) Final temperature of calorimeter + added Calculatio- 69.15 69.22 (g) Mass of cool water in cups, mcw [b- al 22.16 2X.10 (h) Mass of added hot water, myw [c- bl (i) Temperature change of cool water in the calorimeter (ATcw- Tf- Ticw) 19.8 19.3 K K + G) Temperature change of added hot water (ATw= Tf- Tigw 48.S 46.0 K K (k) Calculate the heat lost by the hot water AHEW mwCs ATHW J (I) Calculate the heat gained by the cold water AHCW=mcwCs ATew (m) Calculate the heat gained by the calorimeter AHcal-AHEW-AHew J + J (n) Calculate the calorimeter constant, B J/K JК B-AHcal/ATCW (o) Average calorimeter J/K constant, B Page 96 b0 + + еn Pо

Answer 1

Ans:

First converted given mass in gram to kilogram.

(g) Mass of cool water in cups, m_cw[b-a]

For trial I

69.22/1000=0.06922Kg

For trial II

69.15/1000= 0.06915Kg

(h) Mass of added hot water in cups, m_HW[b-a]

For trial I

28.74/1000= 0.02874Kg

For trial II

28.16/1000=0.02816Kg

Now form standard data table of thermodynamics specific heat capacity of water (C_s) is

4179.6 Jkg⁻¹K⁻¹ ......................... (1)

Note: I assume that specific heat capacity of water is not vary with temperature.

(k) Heat lost by hot water

For trial I

ΔH_HW = m_HWC_sΔT_HW  = 0.02874Kg*4179.6 Jkg⁻¹K⁻¹ * (-48.5K) = - 5825.90J

NOTE: Here minus sign show that heat is lost from hot water.

For trial II

ΔH_HW = m_HWC_sΔT_HW  = 0.02816Kg*4179.6 Jkg⁻¹K⁻¹ * (-46K) = - 5414.09J

NOTE: Here minus sign show that heat is lost from hot water.

(i) Heat gained by cold water

For trial I

ΔH_CW = m_CWC_sΔT_CW =0.06922Kg*4179.6 Jkg⁻¹K⁻¹ * 19.8K = 5728.38J

For trial II

ΔH_CW = m_CWC_sΔT_CW  =0.06915Kg*4179.6 Jkg⁻¹K⁻¹ *19.3K = 5578.07J

(m) Heat gained by Calorimeter

For Trial I

ΔH_cal = -ΔH_HW -ΔH_CW = -(-5825.90J)-5728.38J =97.52J

For Trial II

ΔH_cal = -ΔH_HW -ΔH_CW = -(-5414.09J)-5578.07J =-163.98J

NOTE; In second trial your calorimeter is losing heat. It may mistake in reading taking in trial second boiling water temperature.

(n) Calorimeter constant (B)

For trial I

Calorimeter constant (B) = ΔH_cal/ΔT_CW = 97.52J/19.8K = 4.925JK^-1

For trial II

Calorimeter constant (B) = ΔH_cal/ΔT_CW = -163.98J/19.3K = -8.496JK^-1

(O) Average  Calorimeter constant (B)  

  =[ For trial I  Calorimeter constant (B)+For trial I  Calorimeter constant (B) ] / 2 =[4.925JK^-1 -8.496JK^-1]/2

= -1.786JK^-1

Hence Average  Calorimeter constant (B)  is -1.786JK^-1