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Answer 1 · 2017-08-27T23:26:36

The steam could melt 67.7 g of ice.

The heat $q_{1}$ $q_1$ required to melt a given mass $m_{1}$ $m_1$ of ice is given by the formula

$color(blue)(barul|stackrel(" ")(q_1 = m_1 Δ_text(fus)H)|)$

where $Δ_text(fus)H$ is the enthalpy of fusion of ice.

The heat $q_2$ involved when a given mass $m_2$ of steam condenses is given by the formula

$color(blue)(barul|stackrel(" ")(q_2= "-"m_2 Δ_text(cond)H)|)$

where $Δ_text(cond)H$ is the enthalpy of condensation of steam.

In this problem, $q_1 = q_2$ ,

so

$m_1 Δ_text(fus)H = "-"m_2 Δ_text(cond)H$

and

$m_1 = m_2 × "-"(Δ_text(cond)H)/(Δ_text(fus)H)$

∴ $m_1 = "10.0 g" × "-"("-"2257 color(red)(cancel(color(black)("J/g"))))/(333.3 color(red)(cancel(color(black)("J/g")))) = "67.7 g"$