Procedure was; add 50 mL of distilled water to the calirimeter (styrofoam cups).

Question

Procedure was; add 50 mL of distilled water to the calirimeter (styrofoam cups). And add 5 gram of unknown salt to the calorimeter and stir slowly with the thermometer. Question is; if some heat were transferred from the air or Styrofoam cups would the calculated enthalpy of solution of the unknown salt be too high or too low? Explain. Procedure was; add 50 mL of distilled water to the calirimeter (styrofoam cups). And add 5 gram of unknown salt to the calorimeter and stir slowly with the thermometer. Question is; if some heat were transferred from the air or Styrofoam cups would the calculated enthalpy of solution of the unknown salt be too high or too low? Explain. Question is; if some heat were transferred from the air or Styrofoam cups would the calculated enthalpy of solution of the unknown salt be too high or too low? Explain.

Explanation / Answer

More heat is transferred out of water(surroundings), to system and to air/cup, so temperature change of water is greater.

q = m* C*(delta T). where q = molar Internal energy, m = mass of unknown salt, C = heat capacity of calorimeter and delta T = rise in temperature

So q is greater and enthalpy of solution will also be greater. Therefore, the calculated one is too low.

or The enthalpy would be lower because you are losing the salt. Remember, enthalpy is the total energy in a system. When you take away the energy, enthalpy will decrease.

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