Here given that
density of carbon steel is = 7832 kg/m^3
thickness = 2 mm = 0.002 m
width = 3 cm = 0.03 m
now the mass is given as
[tex]m = \rho * w* t * L[/tex]]
[tex]m = 7832*0.002 * 0.03 * L[/tex]
[tex]m = 0.47*L[/tex]
now we will use the formula of heat
[tex]Q = m s \delta T[/tex]
now plug in all values
[tex]Q = 0.47* L * s* (527 - 127)[/tex]
here s = specific heat capacity of carbon steel
s = 502.4 J/Kg C
[tex]Q = 0.47*L * 502.4 * 400[/tex]
[tex]Q = 94435.1*L[/tex]
now if we find the rate of heat flow
[tex]\frac{dQ}{dt} = 94435.1 * \frac{dL}{dt}[/tex]
[tex]119 * 10^3 = 94435.1 * v[/tex]
[tex]v = 1.26 m/s[/tex]
so its speed is 1.26 m/s