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
Answer:
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
Answer: The x-ray beams are narrowed and shaped with a collimator in the treatment machine head before delivery to the patient.
Explanation:
Answer:
The exit velocity air is 77.56 m/s.
Explanation:
Specific heat of air at 60°C is about 1.006 KJ/kg-K (from standard table).
Now from first law for open system
For ideal gas h=CT
Given that 
°C
°C
By putting the values
So the exit velocity air is 77.56 m/s.
Answer:
(A) 654.545 Kw
(B) 
Explanation:
We have given resistance of the toaster 
Resistance of nichrome heating element 
Both the resistances are connected in series so same current will flow through the circuit
Potential difference across the toaster V = 120 volt
So current 
(a) Power dissipated in toaster 
(B) Power dissipated in heating element 
