(a) The specific heat capacity of copper is 400 \(JKg^{-1}K^{-1}\) means that 400 J of...
(a) The specific heat capacity of copper is 400 \(JKg^{-1}K^{-1}\) means that 400 J of heat energy is required to raise (or lower) the temperature of 1 kg of a piece of copper by 1 kelvin.
(b) Experiment to determine the specific heat capacity of copper using a copper ball. The piece of copper ball is suspended by a thread and placed in a beaker of boiling water. While the temperature of the copper is being raised to boiling point of water, a copper calorimeter is half filled with water. The initial temperature of the water is measured and recorded and the hot copper ball is lifted by the thread and transferred to the calorimeter when the water has reached boiling point. The calorimeter cover is put on, the mixture is well stirred and the final steady temperature read from the themometer. The copper ball is weighed either before or after the experiment. The result is tabulated as shown below.
Mass of copper ball - m (kg)
Temperature of boiling water - \(\theta\) (K)
Mass of copper calorimeter - \(m_{c}\) (kg)
Mass of calorimeter + water - \(m_{x}\) (kg)
Mass of water \((m_{x} - m_{c})\) - \(m_{w}\) (kg)
Specific heat capacity - C \((Jkg^{-1} K^{-1})\)
Initial temperature of water = \(\theta_{w}\) (K)
Final temperature of water = \(\theta_{f}\) (K)
Using Heat lost by copper ball = Heat gained by calorimeter + Heat gained by water
\(mc(\theta - \theta_{f}) = m_{c} c (\theta_{f} - \theta_{w}) + m_{w} c_{w} (\theta_{f} - \theta_{w})\)
\(c = \frac{m_{w} c_{w} (\theta_{f} - \theta_{w})}{m (\theta - \theta_{f}) - m_{c} (\theta_{f} - \theta_{w})}\)
Precautions: (i) The copper ball should be transferred quickly into the calorimeter without splashing. (ii) Just before transfer, the copper should be shaken slightly to remove adhering hot water.
(iii) Heat lost by copper ball = heat gained by copper calorimeter + Hoat gained by water.
\(m_{c} (\theta - \theta_{f}) = m c (\theta_{f} - \theta_{w}) + m_{w} c_{w} (\theta_{f} - \theta_{w})\)
\(20 \times 0.4 \times (200 - \theta_{f}) = 60 \times 0.4 \times (\theta_{f} - 30) + 50 \times 4.2 \times (\theta_{f} - 30)\)
\(8(200 - \theta_{f}) = 24(\theta_{f} - 30) + 210 (\theta_{f} - 30)\)
\(\theta_{f} = \frac{8620}{242} = 35.6K\)
Explanation
(a) Whenever there is a change in the magnetic flux linked with a circuit an electromotive force is induced, the strength of which is proportional to the rate of change of the flux linked with the circuit. Lenz's law states that the direction of the induced current is always such as to oppose the charge producing it.
(b) Lenz's law illustrates the law of conservation of energy. In this case mechanical energy i.e kinetic energy, due to the motion of the bar magnet is changed to electrical energy in the form of induced current.
(c).jpg)
A simple d.c. electric motor can supply direct current (d.c.) by using a split - ring commutator.The brush (-) touches one half of the split ring X while the other brush (+) touches the other half Y. As the coil rotates it cuts through the magnetic flux linking the magnets and current is induced which flows through the coil as shown in the diagram. As the coil rotates past the vertical Y moves round to touch (-) while X moves round to touch (+). This action reverses the connections to the external resistance connected to (-) and (+). The reverse happens when the current is just zero (coil vertical) so instead of the current reversing in the external resistance, it continues to flow in the same direction.
c(ii) (1) Part of the current is wasted in form of heat due to the resistance of the coil and this reduces efficiency;
(2) Friction between the moveable parts of the motor makes the efficiency to reduce.