Question

You drop an ice cube into an insulated bottle full of water and wait for the ice cube to completely melt. The ice cube initially has a mass of 80.0 g and a temperature of 0°C. The water (before the ice cube is added) has a mass of 860 g and an initial temperature of 26.0°C. What is the final temperature (in °C) of the mixture? (Assume no energy is lost to the walls of the bottle, or to the environment.)

Answer 1

`T_(f)` = 17º C

Step-by-step explanation:

This is a calorimetry problem, where heat is yielded by liquid water, this heat is used first to melt all ice, let's look for the necessary heat (Q1)

Let's reduce the magnitudes to the SI system

Ice m = 80.0 g (1 kg / 1000 g) = 0.080 kg

L = 3.33 105 J / kg

Water M = 860 g = 0.860 kg

`c_(e)` = 4186 J / kg ºC

Q₁ = m L

Q₁ = 0.080 3.33 10⁵

Q₁ = 2,664 10⁴ J

Now let's see what this liquid water temperature is when this heat is released

Q = M
`c_(e)` ΔT = M
`c_(e)` (T₀₁ -
`T_(f1)`)

Q₁ = Q

`T_(f1)` = T₀₁ - Q / M ce

`T_(f1)` = 26.0 - 2,664 10⁴ / (0.860 4186)

`T_(f1)` = 26.0 - 7.40

`T_(f1)` = 18.6 ° C

The initial temperature of water that has just melted is T₀₂ = 0ª

The initial temperature of the liquid water is T₀₁= 18.6

m
`c_(e)`
`T_(f)` + M
`c_(e)`
`T_(1)` = M
`c_(e)` T₀₁ - m
`c_(e)` T₀₂o2

`T_(f)` = (M To1 - m To2) / (m + M)

`T_(f)` = (0.860 18.6 - 0.080 0) / (0.080 + 0.860)

`T_(f)` = 17º C

gg