One end of a uniform well-lagged metal bar of length 0.80 m and cross-sectional area of 4.0 x 10-m” is kept in steam at 100 °C while the other end is in melting ice in a well-lagged
container. The ice melts at a steady rate of 5,5 x 10 * kgs and the thermal conductivity of
the material of the bar is 401 Wm''K! Calculate the specific latent heat of fusion of ice.
(3 marks)

Complete question is;

One end of a uniform well-lagged metal bar of length 0.80 m and cross-sectional area of 4.0 x 10-3m2 is kept in steam at 100 ˚C while the other end is in melting ice in a well-lagged container. The ice melts at a steady rate of 5.5 x 10- 4 kgs-1 and the thermal conductivity of the material of the bar is 401 Wm-1K-1. Calculate the specific latent heat of fusion of ice

Specific Latent heat of fusion;

L_f = 13.6 × 10^(5) J/kg

Explanation:

We are given;

Length of bar; L = 0.8 m

Area;A = 4 × 10^(-3) m²

Temperature;ΔT= 100°C = 100 + 273 = 373 K

Rate of melting;m/t = 5.5 × 10^(-4) kg/s

Thermal conductivity;k = 401 W/m·K

Latent heat of fusion has a formula;

ΔQ/Δt = (m/t)•L_f

So, L_f = (ΔQ/Δt)/(m/t) - - - (1)

We also know that ;

ΔQ/Δt = (ΔT × k × A)/L

Plugging in the relevant values, we have;

ΔQ/Δt = (373 × 401 × 4 × 10^(-3))/0.8

ΔQ/Δt = 747.865 J/S

Plugging this value for ΔQ/Δt in equation 1 gives;

L_f = 747.865/(5.5 × 10^(-4))

L_f = 13.6 × 10^(5) J/kg

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y