Question #37c52

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Question #37c52

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Answer 1 · 2016-11-05T14:53:37

$Δ_"soln"H = "-1100 J/g"$

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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Question #37c52

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