# SPH 203,293 CEP CAT ONE Solved

(1) A resistance wire has a resistance value of 3.115 Ω at 27 ^{0}C. At what temperature will the resistance drop to 3 Ω? Take the temperature coefficient of the resistance wire as 4.2×10/^{0}C.

(i)

Since only the temperature coefficient is provided (can assume β=0), equation (i) may be reduced

(ii)

If R_{θ} = R_{1 }at θ=θ_{1}and R_{θ} = R_{2 }at θ=θ_{2} it follows from equation (ii) that;

(iii)

(iv)

Dividing equation (i) by (ii) leads to;

R_{1}=3.115 Ω; θ_{1}=27 ^{0}C; R_{2}=3Ω; α=4.2×10/^{0}C

(2) A galvanometer across a thermocouple reads 0.9 V when the temperature difference between the cold and hot junction is 75 ^{0}C. If the cold junction temperature is increased by 15 ^{0}C, what will be the fractional change in EMF?

If E be the EMF and Δθ the temperature difference, then

Where α is a constant of proportionality

Let E=E_{1 }at Δθ=Δθ_{1 }and E=E_{2 }at Δθ=Δθ_{2}. It follows that;

(i)

(ii)

Dividing (i) with (ii)

E_{1}=0.9 V_{ }at Δθ_{1 }= 75 ^{0}C and E=E_{2 }at Δθ=75-15=60 ^{0}C

(3) A thermally insulated copper wire of length 19 cm and cross-sectional area 0.785 cm^{2} has one end at 100 ^{0}C and the other at 30 ^{0}C. Calculate the amount of heat that will flow in 600 s from the hot to the cold end of the wire. Take thermal conductivity of copper as 380 W/mK.

k=380 W/mK, A=0.785 cm^{2}=785×10^{-7} m^{2}, θ_{2}=30 ^{0}C, θ_{1}=100 ^{0}C, L=19 cm=0.19 m, dt=600s

(4) A glass window pane of total area 0.5 m^{2} and thickness 0.6 cm has a temperature of 23 ^{0}C on the inside and 2 ^{0}C on the outside. Find the amount of heat conducted per hour. Take thermal conductivity of glass as 1.0 W/mk.

k=1.0 W/mK, A=0.5 m^{2}, θ_{2}=2 ^{0}C, θ_{1}=23 ^{0}C, L=0.6 cm=0.006 m, dt=1 h=3600 s

(5)When molten lead metal of mass 3 Kg and temperature 605 K is left to solidify, it is found that the temperature falls from 605 K to 600 K in 10s, remains constant at 600 K for a further 300 s before dropping to 595 K in 8.4 s. Assuming a constant heat loss and taking the specific heat capacity of solid lead as 140 J/kgK, find:

(a) the rate of loss of energy of lead

(b) the specific latent heat of fusion of lead

Heat lost in in 300 s equals product of rate of heat loss and time, ie

(c) the specific heat capacity of molten lead

Heat lost in in 10 s equals product of rate of heat loss and time, ie

## 3 Comment(s)

Dennis Nzuvi katui(Sun, 30th Jul 2023, 4:01 PM)Thankyou Dr. Hope my answer for question 3 which is 6594 won't be marked wrong as I can from the marking scheme the answer is 6.6 × 10^⁴ 6.594 × 10^⁴

Reply

Dennis Nzuvi katui(Sun, 30th Jul 2023, 4:01 PM)Thankyou Dr. Hope my answer for question 3 which is 6594 won't be marked wrong as I can from the marking scheme the answer is 6.6 × 10^⁴ 6.594 × 10^⁴

Reply

Dr. Margaret W. Chege^{Author}(Sun, 30th Jul 2023, 6:00 PM)Answer on the marking scheme is 6.6x10^3 (rounded from 6594. You will not lose marks.

Reply