4. you can 8 liquid water at room temperature and ice in a styrofoaricup uetermi

Question

4. you can 8 liquid water at room temperature and ice in a styrofoaricup uetermine the heat of fusion of ice. In this case, you will deal with usion = the heat flow for the ice as it melts at 0°C to liquid water also at is should be a positive number (the process is endothermic) qf qmeltedice = the heat flow for the melted ice as it warms up from 0°C to the final temperature; this should be a positive number (endothermic) qwater-the heat flow for the room temperature water as it cools to the · final temperature; this should be a negative number (exothermic) In a calorimeter, these three quantities add up to zero, assuming the calorimeter constant is zero. Suppose a 12.500-g sample of ice at 0.00°C is mixed with 100.0 g of liquid water (at 25.00°C) in a calorimeter. The final temperature is found to be 14.36°C. a. Calculate qwater b. Calculate qmelted ice c. Based on qwater and qmelted ice, calculate qrusion d. Calculate qfusion per gram. e. The literature value for the heat of fusion of ice is 334 J/g. Calculate the percent error.

Explanation / Answer

Ans. #a. Heat released when water cools is given by the equation-

q = m s dT                            - equation 1

Where,

q = heat chnage

m = mass of water

s = specific heat of water = 4.184 J g-10C-1

dT = Final temperature – Initial temperature = (T2 – T1)0C

Putting the values in above equation-

            qwater = 100.0 g x (4.184 J g-10C-1) x (14.36 – 25.00)0C = -4451.776 J

The –ve sign of qwater indicates loss of heat from water as it cools

#b. Melted ice is also liquid water. So, qmelted ice can be calculated using equation 1 as follow-

            qmelted ice = 12.500 g x (4.184 J g-10C-1) x (14.36 – 0.0)0C = 751.028 J

#c. While establishing thermal equilibrium, the amount of heat lost by 100.0 g water must be equal to the amount of heat gained by ice while it warms.

While warming, ice first melts at 0.00C (qfusion) and then the melted ice at 0.00C warms to 14.360C (qmelted ice)

So,

            -qwater = qfusion + qmelted ice

            Or, -(-4451.776 J) = qfusion + 751.028 J

            Or, qfusion = 4451.776 J – 751.028 J = 3700.748 J

Therefore, qfusion = 3700.748 J

#d. qfusion per gram = qfusion / Mass of ice = 3700.748 J / 12.500 g = 296.06 J g-1

#e. error in value = Literature value – Calculated value

                                    = 334 J g-1 – 296 J g-1

                                    = 38 J g-1

% error = (error in value / Literature value) x 100

                        = (38 J g-1 / 334 J g-1) x 100

                        = 11.38 %

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