# Question ef8fb

Aug 27, 2017

The steam could melt 67.7 g of ice.

#### Explanation:

The heat ${q}_{1}$ required to melt a given mass ${m}_{1}$ 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)#

${m}_{1} = \text{10.0 g" × "-"("-"2257 color(red)(cancel(color(black)("J/g"))))/(333.3 color(red)(cancel(color(black)("J/g")))) = "67.7 g}$