Question #37c52

1 Answer
Ernest Z.
Nov 5, 2016

#Δ_"soln"H = "-1100 J/g"#

Explanation:

There are two heat transfers involved.

#"heat lost by dissolving NaOH + heat gained by water = 0"#

#q_1 + q_2 = 0#

#m_1Δ_"soln"H + m_2cΔT = 0#

In this problem,

#m_1 = "4.98 g"#

#m_2 = "49.72 g"#
#c = "4.184 J°C"^"-1""g"^"-1"#
#ΔT = T_f - T_i = "50.1 °C - 23.7 °C" = "26.4 °C"#

#q_1 = m_1Δ_text(soln)H= "4.98 g" × Δ_text(soln)H#

#q_2 = m_2cΔT = 49.72 color(red)(cancel(color(black)("g"))) × "4.184 J"·color(red)(cancel(color(black)("°C"^"-1""g"^"-1"))) × 26.4 color(red)(cancel(color(black)("°C"))) = "5492 J"#

#q_1 + q_2 = "4.98 g" × Δ_"soln"H+ "5492 J" = 0#

#Δ_"soln"H = "-5492 J"/"4.98 g" = "-1100 J/g"#

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