PE Integrated Concepts 2 (a) An immersion heater utilizing 120 V can raise the temperature of a 1.00 × 10? -g aluminum cup containing 350 g of water from 20.0ºC to 95.0ºC in 2.00 min. Find its resistance, assuming it is constant during the process. (b) A lower resistance would shorten the heating time. Discuss the practical limits to speeding the heating by lowering the resistance.

Step-by-step solution Given : Voltage = 120 V 2 Mass of the aluminium = 1.00 × 10 -g Mass of the water = 350 g Change in tempreature = 20.0ºC to 95.0ºC Initial tempreature = 20.0ºC Initial tempreature = 95.0ºC Time or duration = 2.00 min Specific heat of Aluminium = 900 J/kg. C 0 Specific heat of water= 4186 J/kg. C 0 Find : Resistance of the Aluminium = Comment on limits to speeding the heating by lowering the resistance. Step 1 of 4 The power is defined as the rate of flow of energy is given by equation (1) P= Q/t…………..(1) Here, Q is defined as the energy and t is the time taken for flow of energy. The electric power is defined as the he rate at which energy is produced through a current carrying element is given by equation (2) as below 2/ P = V R …………….(2) Here, R is the resistance and V is the voltage supplied. Step 2 of 4 Both formulae for power can be compared as,is given by equation (3) as below V /R = Q/t…………….(3) 2 R = V t/ Q ( rearranging the equation (3) we get) The heat produced in the heater is give by equation ( 4) as below Q = m C (T -T) + m C (T - T)..................(4) 1 1 f i 2 2 f i Here, m and m are the masses of aluminium and water respectively .C and C are the 1 2 1 2 specific heat of aluminium and water respectively. T is the final temperature and T is the f i initial temperature of system.