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# Liquid sodium is being considered as an engine coolant- How many grams.docx

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# Liquid sodium is being considered as an engine coolant- How many grams.docx

Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 6.80 MJ of energy (in the form of heat) if the temperature of the sodium is not to increase by more than 10.0 Â°C? Use CP-30.8 J/(K-mol) for Na(I) at 500 K. Number
Solution
Heat absorbed by liquid sodium = specific heat of sodium x mass of sodium x temperature change
We use above formula , where Heat absorbed = 6.8 MJ = 6.8 x 10^ 6 J
dT = change in temperature = 10C = 10K ( since its change in temp)
Cp = 30.8 J/Kmol , = ( 30.8 J/Kmol) x ( 1mol /23g) = 1.339 J/gK
now substituting we get
6.8 x 10^ 6 J = 1.339 J/gK x mass x 10K
mass = 507842 g
.

Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 6.80 MJ of energy (in the form of heat) if the temperature of the sodium is not to increase by more than 10.0 Â°C? Use CP-30.8 J/(K-mol) for Na(I) at 500 K. Number
Solution
Heat absorbed by liquid sodium = specific heat of sodium x mass of sodium x temperature change
We use above formula , where Heat absorbed = 6.8 MJ = 6.8 x 10^ 6 J
dT = change in temperature = 10C = 10K ( since its change in temp)
Cp = 30.8 J/Kmol , = ( 30.8 J/Kmol) x ( 1mol /23g) = 1.339 J/gK
now substituting we get
6.8 x 10^ 6 J = 1.339 J/gK x mass x 10K
mass = 507842 g
.

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### Liquid sodium is being considered as an engine coolant- How many grams.docx

1. 1. Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 6.80 MJ of energy (in the form of heat) if the temperature of the sodium is not to increase by more than 10.0 Â°C? Use CP-30.8 J/(K-mol) for Na(I) at 500 K. Number Solution Heat absorbed by liquid sodium = specific heat of sodium x mass of sodium x temperature change We use above formula , where Heat absorbed = 6.8 MJ = 6.8 x 10^ 6 J dT = change in temperature = 10C = 10K ( since its change in temp) Cp = 30.8 J/Kmol , = ( 30.8 J/Kmol) x ( 1mol /23g) = 1.339 J/gK now substituting we get 6.8 x 10^ 6 J = 1.339 J/gK x mass x 10K mass = 507842 g