Question

Data and Calculations 1. Determining the calorimeter constant Data % 8 (a) Mass of empty Styrofoam cups (b) Mass of cups + 70 ml water (c) Mass of cups + 70 ml water + 30 ml hot water (d) Initial temperature of water in the calorimeter (cups) (e) Temperature of the boiling water bath Trial 1 20.5 86.6 1 113.6 23 95 39 Trial 2 % 20,5 8 87.3 & 114.2 °C 23 C 96 °C 40 °C cc °C (f) Final temperature of calorimeter + added hot water Calculations a (g) Mass of cool water in cups, mcw -b (h) Mass of added hot water, miw 66. 1 8 66.8 8 27 8 26.9 8 + 16 K+ 17 - 54 x-56 (i) Temperature change of cool water in the calorimeter (ATCW - T - Tid 0) Temperature change of added hot water (AT - Tip-Tie (k) Calculate the heat lost by the hot water AH MC, ATI 27 56 (1) Calculate the heat gained by the cold water AHW - McwCATCW (m) Calculate the heat gained by the calorimeter Hal-aww - AHCW (n) Calculate the calorimeter constant. B B-AHJATOW .5.AT XHAUS

Just need a little help figuring out how to make the calculations from my data

Answer 1

Let's pick up where you left off:

The heat lost by the hot water can be calculated as:

$\Delta H_{HW}=m\cdot c\cdot \Delta T=27g\cdot 4.184\frac{J}{g\cdot K}\cdot -56K=-6326J$

$\Delta H_{HW}=m\cdot c\cdot \Delta T=26.9g\cdot 4.184\frac{J}{g\cdot K}\cdot -56K=-6303J$

Where 4.184 J/(gK) is the heat capacity of water.

The heat gained by the cold water can be calculated in the same way:

$\Delta H_{CW}=m\cdot c\cdot \Delta T=66.1g\cdot 4.184\frac{J}{g\cdot K}\cdot 16K=4425J$$\Delta H_{CW}=m\cdot c\cdot \Delta T=66.8g\cdot 4.184\frac{J}{g\cdot K}\cdot 17K=4751J$

The heat gained by the calorimeter is:

$\Delta H_{cal}=6326J-4425J=1901J$

$\Delta H_{cal}=6303J-4751J=1552J$

And the calorimeter constant:

$B=\frac{1901J}{16K}=118.8\frac{J}{K}$

$B=\frac{1552J}{17K}=91.3\frac{J}{K}$

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Mass of water in cups 100 g in both cases (due to the density being 1 g/mL)

Mass of added hot metal: Trial 1: 45.3 g; Trial 2 = 31.3 g

Temperature change of the water: Trial 1: 1.2 K; Trial 2: 1.7 K

Temperature change of the metal: Trial 1: -75.3 K; Trial 2: -74.8 K

heat gained by the cold water:

$\Delta H_{CW}=m\cdot c\cdot \Delta T=100g\cdot 4.184\frac{J}{g\cdot K}\cdot 1.2K=502.1J$$\Delta H_{CW}=m\cdot c\cdot \Delta T=100g\cdot 4.184\frac{J}{g\cdot K}\cdot 1.7K=711.3J$

Heat gained by the calorimeter:

$\Delta H_{cal}=B\cdot \Delta T=118.8\frac{J}{K}\cdot 1.2K=142.6J$

$\Delta H_{cal}=B\cdot \Delta T=91.3\frac{J}{K}\cdot 1.7K=155.2J$

Total heat gained:

$502.1J+142.6J=644.7J$

$711.3J+155.2J=866.4J$

Heat lost by the metal:

$-644.7J$

$-866.4J$

Heat capacity of the metal:

$c=\frac{\Delta H_{metal}}{m\cdot \Delta T}=\frac{-644.7J}{45.3g\cdot -75.3K}=0.189\frac{J}{g\cdot K}$

$c=\frac{\Delta H_{metal}}{m\cdot \Delta T}=\frac{-866.4J}{31.3g\cdot -74.8K}=0.370\frac{J}{g\cdot K}$