If 100mL of 1mol HCL is mixed with 100 mL of 1mol NaOH, a temp rise of 6.2C occurs. What temp change will be obserbed, if under the same conditions, 10mL of 1mol NaOH is mixed with 10mL 1mol HCL?
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Why wouldn't it be the same? One gets 1/10 th of the heat, but has only 1/10th of the solution to heat up?
This ignores thermal loss from surface areas.
To calculate the temperature change, you can use the concept of heat transfer. In this case, the heat released or absorbed is equal to the heat capacity (C) multiplied by the change in temperature (ΔT) and the amount of substance involved (moles).
In the first scenario, 100 mL of 1 M HCl is mixed with 100 mL of 1 M NaOH, resulting in a temperature rise of 6.2 °C.
To find the heat released or absorbed, you can use the equation:
q = C × ΔT
The moles of HCl and NaOH should be the same because they have the same concentration and same volume. So, in this case, 1 mole of each substance is involved.
Therefore, the heat released or absorbed can be calculated as follows:
q = (C × ΔT) × moles
q = (C × 6.2 °C) × 1 mole
Now, in the second scenario, 10 mL of 1 M NaOH is mixed with 10 mL of 1 M HCl. The volume and concentration are both 1/10th of the first scenario.
It is important to note that thermal loss from the surface areas is ignored in this calculation, as you pointed out. This means that the entire heat released or absorbed is taken into account.
Now, to find the expected temperature change, we can use the same equation:
q = (C × ΔT) × moles
In this case, the moles of NaOH and HCl would also be 1/10th of the first scenario as their concentration and volume are reduced by a factor of 1/10.
q = (C × ΔT) × (moles × (1/10))
We know that the moles of NaOH and HCl in the first scenario is 1. Therefore, the moles in the second scenario would be (1/10).
q = (C × ΔT) × (1 × (1/10))
The observed temperature change can be calculated accordingly.
Please note that this calculation considers the ideal case, assuming no heat loss or other factors influencing the temperature change. In real-world scenarios, variations might occur due to factors like heat loss, heat capacity, and other complications.