Before beginning this experiment in the laboratory, you should be able to answer

Question

Before beginning this experiment in the laboratory, you should be able to answer the following questions: degree Define endothermic and exothermic reactions in terms of the sign of AH. A 580-mL sample of water was cooled from 60.0 degree C to 10.0degreeC How much heat was lost? Define the term heat capacity. How many joules are required to change the temperature of 50.0 g of water from 23.3degreeC to 47.6degreeC? Define the term specific heat. Calculate the final temperature when 50 mL of water at 60 degree C are added to 25 mL of water at 25degreeC. Describe how you could determine the specific heat of a metal by using the apparatus and techniques in this experiment. A piece of metal weighing 5.10 g at a temperature of 48.6degreeC was placed in a calorimeter into 20.00 mL of water at 22.1degreeC, and the final equilibrium temperature was found to be 26.8degreeC. What is the specific heat of the metal?

Explanation / Answer

Answer – 1) Endothermic reaction – The reaction in which energy gets absorbed from the system called endothermic reaction. As there is energy gets absorbed, means we need to provide energy like heat the reaction, so the sign of the enthalpy (H) is positive.

Exothermic reaction – The reaction in which energy gets released to the system called exothermic reaction. As there is energy gets released, means energy like in the form of heat the outside the reaction, so the sign of the enthalpy (H) is negative.

2) Given, volume of water = 580 mL , ti = 60.0oC, tf = 10.0oC

We know density of water is 1.0 h/mL , so mass of water = 580 g

We know the formula for calculating the

Heat, q = m*C*t

            = 580 g * 4.184 J/goC * ( 10-60)oC

            = -1.213*105 J

            = -121.3 kJ

So, -121.3 kJ heat was lost.

3) Heat capacity – The heat which need for the temperature increased by one degree is called heat capacity. So heat capacity unit is like J/oC.

4) Given, , mass of water = 50 g , ti = 23.3 oC, tf = 47.6 oC

We know the formula for calculating the

Heat, q = m*C*t

            = 50 g * 4.184 J/goC * ( 47.6 – 23.3)oC

            = 5083.6 J

So, 5083.6 J heat are required to change the 50 g of water from 23.3 oC, to 47.6 oC.

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