The readings obtained in the experiment are given below.mass of aluminium block = 0.930 kginitial temperature of block = 13.1 °Cfinal temperature of block = 41.3 °Celectrical energy supplied = 23 800 J(a) Define specific heat capacity................................................................................................................................................................................................................................................................................................. [2](b) Use the readings above to calculate the specific heat capacity of aluminium.State the equation you use.specific heat capacity = ......................................................... [3]PhysicsAndMathsTutor.com(c) Because the student knows it is good scientific practice to repeat readings, after a short timehe carries out the experiment again, supplying the same quantity of electrical energy.This time the temperature readings are:initial temperature of block = 41.0 °Cfinal temperature of block = 62.1 °C(i) Use these figures to calculate a second value for the specific heat capacity ofaluminium.specific heat capacity = ......................................................... [1](ii) The student did not make any mistakes when taking the readings.Suggest why the second value for the specific heat capacity of the aluminium is greaterthan the first................................................................................................................................................................................................................................................................................. [2](d) Suggest two ways of improving the experiment in order to give as accurate a result aspossible.1. ..................................................................................................................................................................................................................................................................................................2. ............................................................................................................................................................................................................................................................................................. [2]
Question
The readings obtained in the experiment are given below.mass of aluminium block = 0.930 kginitial temperature of block = 13.1 °Cfinal temperature of block = 41.3 °Celectrical energy supplied = 23 800 J(a) Define specific heat capacity................................................................................................................................................................................................................................................................................................. 2 Use the readings above to calculate the specific heat capacity of aluminium.State the equation you use.specific heat capacity = ......................................................... [3]PhysicsAndMathsTutor.com(c) Because the student knows it is good scientific practice to repeat readings, after a short timehe carries out the experiment again, supplying the same quantity of electrical energy.This time the temperature readings are:initial temperature of block = 41.0 °Cfinal temperature of block = 62.1 °C(i) Use these figures to calculate a second value for the specific heat capacity ofaluminium.specific heat capacity = ......................................................... 1 The student did not make any mistakes when taking the readings.Suggest why the second value for the specific heat capacity of the aluminium is greaterthan the first................................................................................................................................................................................................................................................................................. 2 Suggest two ways of improving the experiment in order to give as accurate a result aspossible.1. ..................................................................................................................................................................................................................................................................................................2. ............................................................................................................................................................................................................................................................................................. [2]
Solution
(a) Specific heat capacity is the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius.
(b) The specific heat capacity can be calculated using the formula: Specific heat capacity (c) = Q / (m * ΔT) where Q is the heat energy supplied, m is the mass of the substance, and ΔT is the change in temperature.
Given: Q = 23,800 J m = 0.930 kg ΔT = 41.3°C - 13.1°C = 28.2°C
Substituting these values into the formula gives: c = 23,800 J / (0.930 kg * 28.2°C) = 900 J/kg°C
(c) (i) Using the same formula and the new temperature readings:
ΔT = 62.1°C - 41.0°C = 21.1°C
c = 23,800 J / (0.930 kg * 21.1°C) = 1200 J/kg°C
(ii) The second value for the specific heat capacity of the aluminium is greater than the first because the initial temperature of the block was higher in the second experiment. The specific heat capacity of a material can vary with temperature, and in this case, the aluminium block may have required more energy to increase its temperature when it was already at a higher initial temperature.
(d) Two ways of improving the experiment to give as accurate a result as possible could be:
- Ensuring that the aluminium block is completely isolated during the experiment to prevent heat loss to the surroundings. This could be achieved by using insulating materials around the block.
- Using a more accurate method to measure the temperature change, such as a digital thermometer, to reduce the uncertainty in the temperature readings.
Similar Questions
Calculate the weight in kg of a block of aluminum at 93 °C placed in 3.3 kg of water at 25 °C if the final temperature becomes 33 °C. The heat capacity of water = cw = 4184 J/kg°C, and the heat capacity of Aluminum = cAl = 900 J/kg°CRound answer to the nearest two decimal places
Calculate the energy required to heat 567.0mg of aluminum from 12.5°C to 24.0°C. Assume the specific heat capacity of aluminum under these conditions is ·0.903J·g−1K−1 . Be sure your answer has the correct number of significant digits.
A sample of aluminum, which has a specific heat capacity of 0.897·J·g−1°C−1, is put into a calorimeter (see sketch at right) that contains 100.0g of water. The aluminum sample starts off at 90.6°C and the temperature of the water starts off at 25.0°C. When the temperature of the water stops changing it's 32.3°C. The pressure remains constant at 1atm.Calculate the mass of the aluminum sample. Be sure your answer is rounded to the correct number of significant digits.
If 1.9 kJ of heat is transferred to 96 g aluminum at 113oC, what would the new temperature of the aluminum be? (The specific heat capacity of aluminum is 0.897 J/g•oC.)A.113oCB.114oCC.22oCD.135oC
Aluminum has a specific heat of 0.902 Jg-1K-1. How much heat is lost when a piece ofaluminum with a mass of 23.984 g cools from a temperature of 415.0° C to atemperature of 22.0° C?
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.