00:01
Hello students in this question we have given an aluminum vessel of mass m equals to 0 .5 kg containing mass of water equals to 0 .2 kg at temperature t1 equals to 20 degrees centigrade okay and a block of iron so mass of the iron this is equals to 0 .2 kg initially at a temperature t initial equals to 100 degrees centigrade is gently put into the water so we have to find the equilibrium temperature of the mixture.
00:32
The specific heat capacity for the aluminum, that is c, aluminum, it is given as 910 jule per kilogram kelvin and specific heat for the iron, it is 470 joule per kilogram kelvin and specific heat for the water it is equals to 4200 joule per kilogram kelvin respectively.
00:57
So, we can say here that the from the principle of calorimetry, the heat gained by the heat gained by vessel and water, vessel plus water, this will be equals to heat lost by heat lost by iron, okay? so the equilibrium temperature is capital t, so heat lost by the water and iron will be calculated from here okay so heat gained by the vessel will be mass of the vessel that is aluminium and multiplied by specific heat of aluminium multiplied by temperature difference of t which is equilibrium temperature minus t1 plus heat gained by the water will be mass of the water specific heat of the water and equilibrium temperature of the system minus initial temperature of the water which is this 20 degree centigrade okay, so t minus t1.
02:02
Okay, so we will substitute values later and this is equals to heat lost by the iron.
02:07
So mass of the iron multiplied by specific heat of the iron and temperature difference will be the initial temperature of the iron minus equilibrium temperature t...