if a thermometer reads freezing point of water as 20 degree celcius and boiling point as 150degree celcius , how much thermometer read when the actual temperature is 60 degree celcius?

Approved

Water acquires maximum density at 4 oC. The hydrogen bonds in water keep on forming and breaking. Due to hydrogen bonding, gaps are introduced in the structure of water and various structural units are formed. As the temperature is raised from 0 oC, more hydrogen bonds break than form and thus the structural units collapse. This results in the filling of gaps in the water by smaller structural units or molecules which in turn increases the density of water. As the temperature is further increased the smaller structural units and molecules acquire speed and move faster which causes water to expand and thus reduce the density of water. Below 4 oC the effect of breaking of hydrogen bonds overcomes the increased speed of the smaller structural units and molecules and thus the density of water increases till 4 oC. But after this temperature the speed of units and molecules overcome the effect of breaking hydrogen bonds and thus density of water is decreased. Hence, water has a maximum density at 4 oC.

The relation between the change in heat energy and the change in temperature of a metal of mass m having specific heat capacity equals to S is given as, ? Q = m c ? T The above equation can also be written as, d Q = m S d T On susbtituting for S and unit mass (m=1 kg) we get, d Q = a T 3 d T Now integrating both sides with the range 0 to Q for heat energy and from 1 K to 2 K for the temperature we get, ? 0 Q d Q = a ? 1 2 T 3 d T Q 0 Q = a T 4 4 1 2 Q = a 2 4 - 1 4 4 Q = a 16 - 1 4 Q = 15 4 a

Let us consider a small element of mass dm in the ring. Let it subtend elementary angle d? at the centre. Due to rotation the mass dm will experience a centrifugal force F outwards. Due to this forces F’ develop within the ring to keep it in equilibrium. Let A be the CSA of the wire. From equilibrium of three forces we have F/sin(180 – d?) = F’/sin(90 + d?) => F/sin(d?) = F’/cos(d?) As d? is very small, sin (d?) = d? and cos (d?) = 1 => F/ d? = F’ => F = F’ d? ---1. Now F = (dm)?2r dm = dAdl Again, dl = rd? Thus, F = dA?2r2d? From 1. F’ d? = dA?2r2d? => F’ = dA?2r2 => F’/A = d?2r2 Now the maximum stress the ring can withstand is s => s = d?2r2 => ? = 1/rv(s/d) Let f be number of rotation per second => ? = 2pf = 1/rv(s/d) => f = 1/2prv(s/d)

-1.13 KJ

the answer is

the given reaction

GIVE ME PROCEDURE HOW TO GET SOLUTION.