The ΔH for the solution process when solid sodium hydroxide dissolves in water is 44.4 kJ/mol. When a 10.1-g sample of NaOH dissolves in 250.0 g of water in a coffee-cup calorimeter, the temperature increases from 23.0 °C to __________°C. Assume that the solution has the same specific heat as liquid water, i.e., 4.18 J/g-K.

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The heat absorbed by the water is:$$q = mC_p \Delta T = (250.0 	ext{ g})(4.18 	ext{ J/g-K})(T_f - 23.0 \degree C)$$The heat released by the NaOH is:$$q = - \Delta H_{soln}n = -(44.4 	ext{ kJ/mol})(10.1 	ext{ g NaOH})(1 	ext{ mol NaOH}/40.00 	ext{ g NaOH}) = -11.2 	ext{ kJ}$$Setting these two heats equal to each other, we get:$$(250.0 	ext{ g})(4.18 	ext{ J/g-K})(T_f - 23.0 \degree C) = -11.2 	ext{ kJ}$$Solving for $T_f$, we get:$$T_f = 33.3 \degree C$$

The ΔH for the Solution Process When Solid Sodium Hydroxide