A steel container of mass 160 g contains 26.0 g of ammonia, NH2, which has a molar mass of 17.0 g/mol. The container and gasare in equilibrium at 19.0°C. How much heat (in J) has to be removed to reach a temperature ofvolume of the steel. (The specific heat of steel is 452 J/(kg °C). Enter the magnitude.)20.0°C? Ignore the change in

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Asked Jul 5, 2019
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A steel container of mass 160 g contains 26.0 g of ammonia, NH2, which has a molar mass of 17.0 g/mol. The container and gas
are in equilibrium at 19.0°C. How much heat (in J) has to be removed to reach a temperature of
volume of the steel. (The specific heat of steel is 452 J/(kg °C). Enter the magnitude.)
20.0°C? Ignore the change in
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A steel container of mass 160 g contains 26.0 g of ammonia, NH2, which has a molar mass of 17.0 g/mol. The container and gas are in equilibrium at 19.0°C. How much heat (in J) has to be removed to reach a temperature of volume of the steel. (The specific heat of steel is 452 J/(kg °C). Enter the magnitude.) 20.0°C? Ignore the change in

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Expert Answer

Step 1

Given:

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Mass of steel container, m160 g 160 x10 kg Mass of Ammonia,m = 26.0g = 26.0 x 103 kg Molar mass of ammonia,=17.0 g/mol=17.0 x 10 kg/mol Specific heat of steel, C 452 J/kg.K Initial temperature, T 19.0°C Final temperature, T =-20.0°C

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Step 2

Calculation:

The amount of heat for ammonia is given by,

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Q =nC,AT Where n Number of moles C, Specific heat AT Differencein temperature

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Step 3

The amount of heat for s...

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Q mC,AT Where Mass of steel m CSpecific heat of steel AT Differencein temperature

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