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“what amount of water would be needed to be added to 10g of ammonium nitrate to change the temperature of an instant cold pack form 20 degrees to 5 degrees? (6 marks)
8 years ago
Chemistry Question asked by Wewik
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1 Answer
You can calculate the actual answer by the following method:
>>>1. Assume: the change is adiabatic (i.e. very fast or very well insulated the specific heat capacity of the solution is the same as for water (c=4.2kJ/kg/K) that the Enthalpy ("Heat") of Solution for NH4NO3 is +25.8kJ/mol.
>>>2. Find the molar mass of NH4O3: (14+4x1+14+3x16 = 80)
>>>3. Find the no. of moles you have: 10/80 = 1/8
>>>4. Calculate the Change in Heat (dH) as dH = 25.8kJ/mol x 1/8mol = 3.225kJ
>>>5. Realise that dH is also: dH = Ms x c x dT [where Ms is the mass of the solution, c is the specific heat capacity given above and dT is the temperature change (20 - 5 = 15 in your case).
>>>6. Note that Ms = M + Mw (where M is your "10g" or 0.01kg of NH4NO3 and Mw is the mass of water that you are trying to find)...
>>>So: Ms = dH/c.dT
>>>putting the numbers in gives:
Ms = 3.225/(4.2x15 = 0.05kg
>>>>so finally, Mw = Ms - M = 0.05 - 0.01 = 0.04kg or 40g (i.e. 40ml) of water .... the question vaguely asked for "amount" of water so I guess either the mass or volume would do.
>>>1. Assume: the change is adiabatic (i.e. very fast or very well insulated the specific heat capacity of the solution is the same as for water (c=4.2kJ/kg/K) that the Enthalpy ("Heat") of Solution for NH4NO3 is +25.8kJ/mol.
>>>2. Find the molar mass of NH4O3: (14+4x1+14+3x16 = 80)
>>>3. Find the no. of moles you have: 10/80 = 1/8
>>>4. Calculate the Change in Heat (dH) as dH = 25.8kJ/mol x 1/8mol = 3.225kJ
>>>5. Realise that dH is also: dH = Ms x c x dT [where Ms is the mass of the solution, c is the specific heat capacity given above and dT is the temperature change (20 - 5 = 15 in your case).
>>>6. Note that Ms = M + Mw (where M is your "10g" or 0.01kg of NH4NO3 and Mw is the mass of water that you are trying to find)...
>>>So: Ms = dH/c.dT
>>>putting the numbers in gives:
Ms = 3.225/(4.2x15 = 0.05kg
>>>>so finally, Mw = Ms - M = 0.05 - 0.01 = 0.04kg or 40g (i.e. 40ml) of water .... the question vaguely asked for "amount" of water so I guess either the mass or volume would do.
