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3 years ago

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Two wires have the same resistance and radius. If the wires are made of copper and platinum with resistivities respectively of 1

.70 ✕ 10−8 Ω · m and 11.0 ✕ 10−8 Ω · m, determine the ratio of their lengths.

1 answer:

Zolol [24]3 years ago

3 0

Answer:

L1/L2 = 6.47

Explanation:

In order to calculate the ratio of the lengths of the wires you use the following formula for the resistivity of a wire:

$\rho=\frac{\pi r R}{L}$ (1)

r: radius of the cross-sectional area of the wire

R: resistance of the wire

L: length of the wire

Then, you have for each wire:

$\rho_1=\frac{\pi r_1^2R_1}{L_1}=1.70*10^{-8}\Omega.m\\\\\rho_2=\frac{\pi r_2^2R_2}{L_2}=11.0*10^{10^{-8}}\Omega.m$

The resistance and radius of the wires are the same, that is, R1 = R2 = R and r1 = r2 = r. By taking into account this last and dive the equation for the wire 2 into the wire 1, you obtain:

$\frac{\rho_2}{\rho_1}=\frac{11.0*10^{-8}\Omega.m}{1.70*10^{-8}\Omega.m}=\frac{L_1}{L_2}\\\\\frac{L_1}{L_2}=6.47$

The ratio of the lengthd of the wires is L1/L2 = 6.47

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Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

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Answer:

115 ⁰C

Explanation:

<u>Step 1:</u> The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

$q_{1} +q_{2} =-q_{3}$ -----eqution 1

where,

$q_{1}$ is the heat absorbed by the solid at 0⁰C

$q_{2}$ is the heat absorbed by the liquid at 0⁰C

$q_{3}$ the heat lost by the warmer water sample

Important equations to be used in solving this problem

$q=m *c*\delta {T}$ , where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

$\delta {T}$ is change in temperature

Again,

$q=n*\delta {_f_u_s}$ -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

<u>Step 2:</u> calculate how many moles of water you have in the 100.0-g sample

$=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O$

<u>Step 3: </u>calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

$q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ$

This means that equation (1) becomes

79.13 KJ + $q_{2} = -q_{3}$

<u>Step 4:</u> calculate the final temperature of the water

$79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}$

Substitute in the values; we will have,

$79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})$

79.13 kJ + 990.66J* $(T_{f}-218})$ = -1463J* $(T_{f}-100})$

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* $(T_{f}-218})$ = -1.463KJ* $(T_{f}-100})$

$79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3$

collect like terms,

2.45366 $T_{f}$ = 283.133

∴ $T_{f} =$ = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

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