Please answer questions 4, 5 and 6 This lab is the Enthalpy of Fusion of Ice. It

Question

Please answer questions 4, 5 and 6

This lab is the Enthalpy of Fusion of Ice. It seems to be a rite of passage to calculate how ice melts.

You will use foam cups to insulate and contain the ice water. You'll be measuring the heat absorbed, not the "cold" given off.
Use the equation Q=mc?T to calculate the gain or loss of heat of water and melted ice. (please show your work)

You will use foam cups to insulate and contain the ice water. You'll be measuring the heat absorbed, not the "cold" given off.

Use the equation Q=mc?T to calculate the gain or loss of heat of water and melted ice.

1. Data Table 1: Mass data

58.64

2. Data Table 2: Temperatures

3. Data Table 3: Report Q

4.
Use Equation Qf=mLf and the data in your tables to calculate the heat of fusion of ice.

5. The accepted value for the enthalpy of fusion of ice is 334 J/g. Calculate your error.

6. ?H is normally given in units of kJ/mole. Use your value of Qf and grams/mole of water to find the experimental value ?H for the fusion of ice.

Mass (g) all digits are significant Mass of empty calorimeter (stacked cups) 2.21 Mass of water in calorimeter 53.45 Total mass (calorimeter + water + ice) 51.24 Mass of ice and water (calculated)

58.64

Mass of ice only (calculated) 2.98

Explanation / Answer

% error = | measured - accepted| / |accepted| x 100%.. note the absolute value bars!!!

% error = | 419 J/g - 334 J/g| / |334 J/g| x 100% = 25%... (2 sig figs)

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it works out the same right?..
(a - b) / b = (axc - bxc) / (bxc) doesn't it?

and to convert from J/g to J/mole.. we just multiply everything by (18g/mole).... that's our "c" :)

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but if you insist...
419 J/g x (18.02g / mole) = 7550 J/mole
334 J/g x (18.02g / mole) = 6019 J/mole
then..
% error = | 7550 J/g - 6019 J/g| / |6019 J/g| x 100% = 25%... (2 sig figs)

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NO.. not 125%..

this is 125% error..
125% = |measured - 334| / 334
measured = 752 J/g

use the formula!.. % error = |measured - accepted| / accepted x 100%

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btw.. given that you're calculating % error in your class... you may be asked at another time to calculate % difference. That equation is

% difference = [ | value1 - value2 | / ((| value1 - value2 |) / 2) ] x 100%
again.. | | is absolute value..

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yes.. that's what the calorie meter constant is for... heat absorbed by the calorimeter.

if you think of this equation
Q = m Cp dT
and group like this
Q = (m Cp) dT
and look at the units on m x Cp
g x (J/g

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