How much heat is needed to warm 135 g of water (about 1 cup) from 25 degree C (a

Question

How much heat is needed to warm 135 g of water (about 1 cup) from 25 degree C (about room temperature) to 80 degree C (near its boiling point)? (b) What is the molar heat capacity of water? A piece of zinc weighing 45.8 g was heated from 20.00degree C to 38.00degree C. How much heat was required? The specific heat of zinc is 0.388 J/(gdegree C). When a student mixes 60 mL of 1.0 M HCl and 60 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0 to 34.5 degree C. Calculate the enthalpy change for the reaction in kJ/mol HCl, assuming that the calorimeter loses only a negligible quantity of heat, that the total volume of the solution is 100 mL. that its density is 1.0 g/mL, and that its specific heat is 4.18 J/g-K.

Explanation / Answer

Answer:

1) a) Given Data:

Inlet mass flow of water, m = 135 g = 0.135 kg

Initial temperature, T1 = 25 ° C = 298 K

Final temperature, T2 = 80 ° C = 353 K

Heat required, Q = m * Cp * (T2 - T1)

Mass heat capacity of water, Cp = 4186 J/kg. K

Heat needed, Q = 0.135 kg * 4186 J/kg. K * (353 - 298) K = 31081 J = 31.081 kJ

b) Molar heat capacity of water:

Cpmolar = mass heat capacity * molar mass = 4186 J/kg. K * 0.018 kg/mol (since molar mass of water is 18 kg/kmol)

Cpmolar = 75.348 J/mol. K

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