Question

Suppose you have 600.0 grams of room temperature water (20.0 degrees Celsius) in a thermos. You drop 90.0 grams of ice at 0.00 degrees Celsius into the thermos and shut the lid. (a) What is the equilibrium temperature of the system? (b) How much ice is left (in grams)?

Answer 1

`T_(f)` = 7.02 ° C

Step-by-step explanation:

The liquid water gives heat to melt the ice (Q₁) maintaining the temperature of 0 ° C and then the two waters are equilibrated to a final temperature.

Let's start by calculating the heat needed to melt the ice

Q₁ = m L

Q₁ = 0.090 3.33 10⁵

Q₁ = 2997 10⁴ J

This is the heat needed to melt all the ice

Now let's calculate at what temperature the water reaches when it releases this heat

Q = M
`c_(e)` (T₀ -
`T_(f)`)

Q₁ = Q

`T_(f)` = T₀ - Q₁ / M
`c_(e)`

`T_(f)` = 20.0 - 2997 104 / (0.600 4186)

`T_(f)`= 20.0 - 11.93

`T_(f)` = 8.07 ° C

This is the temperature of the water when all the ice is melted

Now the two bodies of water exchange heat until they reach an equilibrium temperature

Temperatures are

Water of greater mass T₀₂ = 8.07ºC

Melted ice T₀₁ = 0ºC

M
`c_(e)` (T₀₂ -
`T_(f)`) = m
`c_(e)` (
`T_(f)` - T₀₁)

M T₀₂ + m T₀₁ = m
`T_(f)`+ M
`T_(f)`

`T_(f)`= (M T₀₂ + 0) / (m + M)

`T_(f)` = M / (m + M) T₀₂

let's calculate

`T_(f)` = 0.600 / (0.600 + 0.090) 8.07

`T_(f)` = 7.02 ° C