A student determined the calorimeter constant of the calorimeter. The student ad

Question

A student determined the calorimeter constant of the calorimeter. The student added 50.00 ml of cold water to 50.00 ml of heated distilled water in a Styrofoam cup the initial temperature pf the cold water was 21.00 degrees Celsius and of the hot water 29.15 degrees Celsius the maximum temp of the mixture was found to be 24.891 degrees C the maximum temp of the mixture was found to be 24.81 Celsius assume the density of water is 1.00 g ml-1 and the specific heat is 4.184 J g-1 K-1.

1) Determine the delta T for the hot water and the cold water;

2) calculate the heat lost by the hot water;

3)calculate the heat gained by the cold water;

4)Calculate the calorimeter constant, using the delta T of the cold water.

Explanation / Answer

1. delta T for Hot water = 29.15 - 24.891 = 4.259 deg C;

                for cold water = 24.891 - 21 = 3.891 deg C;

2.Heat lost by hot water = mCdt ;

= d * v * C * ( 29.15 - 24.891);

= 1 * 50 *4.184 * 4.259;

= 890.1828;

3.heat gained by the cold water = m *C * dt

dt = 24.891 -21;

m = d * v = 1* 50 = 50 ;

and C =4.184;

heat gained = 50 * 4.184 * 3.891 = 813.9972;

4. Calorimeter constant using dT for cold water ;

C = delta H / dT;

C = 890.1828 / 4.29 = 207.50

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