aBetterChemText: Another Two Character Problem

Monday, July 8, 2019

Another Two Character Problem

A 21.3 g metal spoon (c spoon = 0.074cal/g ^oC) at 82.9^oC is dropped into 280.0 g of water at 11.0^oC. What is the final temperature of the spoon?

This problem involves one item (the water) giving heat to the other (the spoon).

So, the two amounts of heat will be the same, but one will be negative, leading to the formula

$(-1)q_{spoon}=q_{water}$

And since both are temperature changes, we can change that formula to

$(-1)m_{spoon}\cdot c_{spoon}\cdot \Delta T_{spoon}=m_{water}\cdot c_{water}\cdot \Delta T_{water}$

Now, what do we know?

We can't find the ΔT for either the spoon or the water (since we don't have the final T), but we can substitute in the formula for ΔT:

$(-1)m_{s}\cdot c_{s}\cdot (T_{final}-T_{initial ~for~spoon})=m_{w}\cdot c_{w}\cdot (T_{final}-T_{initial ~for~water})$

Now we can do some algebra. First, we'll distribute:

Next, we'll move the T_final factors to the right side and the initial temperature factors to the left.

Then we can isolate T_final

and then sub in our values and solve:

No comments:

Post a Comment

Subscribe to: Post Comments (Atom)