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antiseptic1488 [7]
3 years ago
10

1. A copper block of volume 1 L is heat treated at 500ºC and now cooled in a 200-L oil bath initially at 20◦C. Assuming no heat

transfer with the surroundings, what is the final temperature?
Engineering
1 answer:
MaRussiya [10]3 years ago
7 0

Answer:

final temperature T = 24.84ºC

Explanation:

given data

copper volume = 1 L

temperature t1 = 500ºC

oil volume = 200 L

temperature t2 = 20ºC

solution

Density of copper \rho cu = 8940 Kg/m³

Density of light oil  \rho oil = 889 Kg/m³

Specific heat capacity of copper Cv = 0.384  KJ/Kg.K

Specific heat capacity of light oil Cv = 1.880 KJ/kg.K

so fist we get here mass of oil and copper that is

mass = density × volume   ................1

mass of copper = 8940 × 1 ×  10^{-3}  = 8.94 kg  

mass of oil = 889 × 200 × 10^{-3}  =  177.8 kg  

so we apply here now energy balance equation that is

M(cu)\times Cv \times (T-T1)_{cu} + M(oil) \times Cv\times (T-T2)_{oil}  = 0

put here value and we get T2

8.94\times 0.384 \times (T-500) + 177.8 \times 1.890\times (T-20)  = 0

solve it we get

T = 24.84ºC

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Explanation:

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\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m

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h=\frac{P_2-P_1}{\rho g}+z_2-z_1\\\Righarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+3\\\Righarrow h=13.19\ m\\

Work done by pump

W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 13.193\\\Rightarrow W_{p}=20708.8\ W

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