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Calculate the temperature increase when \( 400 \text{ J} \) of heat is applied to \( 46.8 \text{ g} \) of \(\ce{NaCl}.\)

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**Details and assumptions:**

- The molar heat capacity for \(\ce{NaCl}\) is \( C_p = 50 \text{ J} \cdot \text{mol}^{–1} \cdot \text{K}^{–1} .\)
- The formula weight of \(\ce{NaCl}\) is \( 58.5 \text{ g/mol}. \)

Initially, there are two thermally isolated rooms \( A \) and \( B.\) The heat capacitance of room \( A\) is \( 80 \) quanta of energy and room \( A\) contains \( 56 \) quanta of energy. The heat capacitance of room \( B \) is \( 80 \) quanta of energy and room \( B \) contains \( 24 \) quanta of energy.

**Question**: When the two rooms are thermally connected and allowed to approach equilibrium, how many quanta of energy are transferred, on average?

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