Given the following information, calculate the heat of solution for the salt in the unit of KJ. The calorimeter is prepared with 44.96 g of water, both temperatures are at 22.6 degree Celsius. 2.42 g of salt are dissolved in the water and the temperature falls to 18.1 degree Celsius. (Note: the heat capacity of this calorimeter is 30.1 J/degree Celsius.
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Heat = Heat of the water + Heat of the calorimeter
Q = m x dT x SH + HC x dT (m = mass, dT = temp change, SH = sp. ht. capacity of H2O, HC = heat capacity of calorimeter)
Q = 44.96 g x 4.5 C x 4.184 J/gC + 30.1 J/C x 4.5 C
Q = 847 J + 135 J = 982 J
982 J of heat is absorbed from the surroundings for 2.42 g of salt.
Heat / g = 0.982 kJ / 2.42 g = 0.406 kJ/g of salt
Q = m x dT x SH + HC x dT (m = mass, dT = temp change, SH = sp. ht. capacity of H2O, HC = heat capacity of calorimeter)
Q = 44.96 g x 4.5 C x 4.184 J/gC + 30.1 J/C x 4.5 C
Q = 847 J + 135 J = 982 J
982 J of heat is absorbed from the surroundings for 2.42 g of salt.
Heat / g = 0.982 kJ / 2.42 g = 0.406 kJ/g of salt
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∆H = ∆T x (mass solution x specific heat water + heat capacity calorimeter).....or is the heat capacity of the calorimeter the sum of the cup and the water already combined?
∆H = 4.5ºC x (44.96 x 4.184J/g-ºC + 30.1J/ºC) = 25479.86J or 2.54x10^4kJ
if the heat capacity of the calorimeter already includes the heat capqcity of the water, then:
∆H = ∆T x heat capacity calorimeter = 4.5ºC x 30.1J/ºC = 135.45J or 0.135kJ
∆H = 4.5ºC x (44.96 x 4.184J/g-ºC + 30.1J/ºC) = 25479.86J or 2.54x10^4kJ
if the heat capacity of the calorimeter already includes the heat capqcity of the water, then:
∆H = ∆T x heat capacity calorimeter = 4.5ºC x 30.1J/ºC = 135.45J or 0.135kJ