Question #ef8fb

1 Answer
Aug 27, 2017

Answer:

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