An aluminum rod is 24.0 cm long at 20°C?

An aluminum rod is 24.0 cm long at 20°C and has a mass of 350 gIf 14000 J of energy is added to the rod by heat, what is the change in length of the rod?1 mm

There are two constants you have not supplied: the linear coefficient of thermal expansion for aluminum, and the specific heat of aluminumIs the 1 mm the presumed answer? Is it rounded? I will use the following, obtained from Wikipedia: coefficient of linear thermal expansion α_L for Al: 23 × 10^(-6)/C° specific heat c_p of Al: 0.897 J/g-C° Equation: ΔL L0(α_L)(T2 ? T1))T2 - T1 14000 J /( 350g 0.897 J/g-C°) ≈ 44.6 C° Therefore, ΔL (24.0)(2.3e-5)(44.6) ≈ 0.025cm 0.25mm I get only a quarter of the 1 mm you've (possibly) indicated.

any plant which excavates is called either a digger/backhoe loader or an excavator,, some have tracks, some have rubber tyres, some of the really small ones just drag themselves about

I agree with the first answer by Mam but next time it's on the tip of your tongue do a Google Image search and usually what you are trying to get a pic of one will show just like a digger and the back hoe will show.

Construction Trucks Names

Construction Trucks Names

any plant which excavates is called either a digger/backhoe loader or an excavator,, some have tracks, some have rubber tyres, some of the really small ones just drag themselves about

I agree with the first answer by Mam but next time it's on the tip of your tongue do a Google Image search and usually what you are trying to get a pic of one will show just like a digger and the back hoe will show.

There are two constants you have not supplied: the linear coefficient of thermal expansion for aluminum, and the specific heat of aluminumIs the 1 mm the presumed answer? Is it rounded? I will use the following, obtained from Wikipedia: coefficient of linear thermal expansion α_L for Al: 23 × 10^(-6)/C° specific heat c_p of Al: 0.897 J/g-C° Equation: ΔL L0(α_L)(T2 ? T1))T2 - T1 14000 J /( 350g 0.897 J/g-C°) ≈ 44.6 C° Therefore, ΔL (24.0)(2.3e-5)(44.6) ≈ 0.025cm 0.25mm I get only a quarter of the 1 mm you've (possibly) indicated.