1. A steel railroad track has a length of 30.000m when the temperature is 0°C. What is the length on a hot day when the temperature is 40.0°C. The linear coefficient of steel is 11x10^ -6/ c°.
2. A circular copper ring at 20°c has a hole with an area of 9.980cm². What minimum temperature must it have so that it can be slipped onto a steel having a cross sectional area of 10.000cm². The linear coefficient of copper is 17x10^ -6/ c °.
3. A steal strut near a ships furnace is 2.00m long with a mass of 1.57kg and cross sectional area of 1.00x10^ -4m². During operation of the furnace, the strut absorbs a net thermal energy of 2.50x10^5 J. Find the change in temperature of the strut and also find the increase in length of the strut. The coefficient of steel is 11x10^ -6/ C° and the specific heat is 4485/kg.c°
Please help me answers this with solution. I really don't know how to solve this. Thank you :) Your help is much appreciated.
2. A circular copper ring at 20°c has a hole with an area of 9.980cm². What minimum temperature must it have so that it can be slipped onto a steel having a cross sectional area of 10.000cm². The linear coefficient of copper is 17x10^ -6/ c °.
3. A steal strut near a ships furnace is 2.00m long with a mass of 1.57kg and cross sectional area of 1.00x10^ -4m². During operation of the furnace, the strut absorbs a net thermal energy of 2.50x10^5 J. Find the change in temperature of the strut and also find the increase in length of the strut. The coefficient of steel is 11x10^ -6/ C° and the specific heat is 4485/kg.c°
Please help me answers this with solution. I really don't know how to solve this. Thank you :) Your help is much appreciated.
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1>l=30(1+11*10^-6*40)=..................
2>alpha=17*10^-6
Bita=co ef for area expansion=2*alpha
initial area=a
a(1+2*17*10^-6)=9.980
a=......
say temp=t
10=a*(1+2*17*10^-6*t)
solve it ur self
3>Q=m*s*t
2.5*10^5=1.57*4485*t
t=......
change in length=2*111*10^-6*t (as no initial temp is givn)
2>alpha=17*10^-6
Bita=co ef for area expansion=2*alpha
initial area=a
a(1+2*17*10^-6)=9.980
a=......
say temp=t
10=a*(1+2*17*10^-6*t)
solve it ur self
3>Q=m*s*t
2.5*10^5=1.57*4485*t
t=......
change in length=2*111*10^-6*t (as no initial temp is givn)