Question

Question #ef8fb

Answer 1

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"`