A student is given the task of determining the specific heat capacity of a chunk

Question

A student is given the task of determining the specific heat capacity of a chunk of metal of unknown composition. The following materials are available. According to the laws of thermodynamics, the heat lost by one substance is exactly equal to the heat gained by the other substance in a closed, two-substance system. Explain how. on a microscopic level, thermal energy (heal) is transferred from a hot chunk of metal to liquid water (at a lower temperature than the metal) in a confined space. From an experimental design standpoint, explain how the student can heat the metal chunk to in observable/measurable (with a thermometer) temperature? Explain. The student measures the temperature of 35.00 g of distilled water. 24.2degreeC. The student then places a 2.33 g chunk of metal, heated to 100.2 degreeC. into the 35.00 g of distilled water. What should the student use to contain the metal/water mixture? Justify your answer. At some point following immersion of the hot metal into the room temperature water, a thermal equilibrium between the metal and water is established. How does the student determine the point at which thermal equilibrium is established? The student stirred the mixture as it approached its maximum temperature. Why is this good idea, from an experimental design perspective? Calculate the experimental value for the specific heat capacity of the unknown metal assuming the maximum temperature of the metal-water mixture was found to be 44.5 degree C. C water = 4.18 J/g X °C. The students teacher informs him/her that the actual specific heat capacity of the metal is 24.3 J/g X °C. Provide one reason to account for the difference between the students experimental value and actual value. Justify your answer.

Explanation / Answer

Solution :-

Part a i) hot chuck have higher temperature than the water at room temperature therefore when the hot chunk is added to the water then metal chunk loss the heat and water will absorb it in the form of the radiation.

ii) to observe the temperature of the heated metal chunk student need to heat the metal shuck in the hot waterbath so that the temperature of the water bath can be measured on the thermometer which have same temperature for the metal chunk

part b given data

mass of water = 35.00 g , initial temperature of distilled water = 24.2 C

mass of metal chunk = 2.33 g , temperature of metal chunk = 100.2 C

i)Student need to use the calorimeter to contain the water and metal chunk so that heat cannot be loss to the surrounding.

ii) After the immersion of the hot metal chunk in the room temperature water the temperature of the water will rise gradually and it will remain constant for some time and then again start to drop

the constant temperature indicates that the metal chunk and water are at equilibrium temperature.

iii) stirring is important because it causes the water to circulate and attain the equilibrium temperature thorough out the water.

Part C

Equilibrium final temperature is 44.5 C

Lets first calculate the amount of heat absorbed by water

q= m*c*delta T

= 35.00 g * 4.18 J per gC * (44.5 C – 24.2 C)

= 2969.89 J

Now the amount of heat loss by metal chunk is same as amount of heat absorbed by water

So using this amount of heat lets calculate the specific heat of the metal chunk

c chunk = q/ m*delta T

              = 2969.89 J / 2.33 g * (100.2 C – 44.5 C)

             = 22.88 J per g C

So the specific heat of the metal chunk is 22.88 J/ g C

Part d

actual specific heat = 24.3 J / g C

calculated specific heat = 22.88 J/g C

The calculated specific heat is lower than the actual specific heat therefore the reason behind this would be some of the heat may loss to the surrounding because of imperfect insulation. of the calorimeter.

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