aBetterChemText: A 3rd Two-Character Problem

Monday, July 8, 2019

A 3rd Two-Character Problem

A 67.9 g piece of metal (c=0.069 cal/g ^oC) is heated to 174^oC and then dropped onto a large block of ice at 0^oC. Assuming that the metal cools down to 0^oC, what mass of ice will be melted? (ΔH_fus of ice = 80.0 cal/g)

This problem involves one thing giving away heat (the metal cooling down) and one thing absorbing heat (the ice melting). So the two amounts of heat will be the same leading to the formula

$(-1)q_{metal} = q_{ice}$

The metal is undergoing a change in temperature so the metal side of the equation is ,

$q_{metal} = m \cdot c \cdot \Delta T$

while the ice is undergoing a phase change , so that the ice side of the equation is

$q_{ice} = m \cdot \Delta H_{fus}$

Putting those together gives us the equation:

$(-1)m_{metal}\cdot c_{metal} \cdot \Delta T_{metal} = m_{ice} \cdot \Delta H_{fus}$

Now, let’s see what we know.

Let's rearrange that equation to solve for mass.

$\frac{(-1)m_{metal}\cdot c_{metal} \cdot \Delta T_{metal}}{\Delta H_{fus}} = m_{ice}$

Now, we can substitute in our values and solve.

$\frac{(-1)67.9 ~g\cdot 0.069 \frac{cal}{g \cdot^oC} \cdot 174^oC}{79.7 \frac{cal}{g}} = m_{ice} =10.2 ~g$

No comments:

Post a Comment

Subscribe to: Post Comments (Atom)