A 35-g block of ice at -14°C is dropped into a calorimeter (of negligible heat capacity) containing 400 g of water at 0°C. When the system reaches equilibrium, how much ice is left in the calorimeter? The specific heat of ice is 2090 J(kg K) and the latent heat of fusion of water is 33.5 x 10^4 J/kg.

well, we have .035 kg of ice at -14

we have .4 kg of water at 0

If those really are the numbers,
the answer is that some of the water gets frozen by the ice which is 14 degrees BELOW freezing. At least 35 grams of ice will be left.
How much water can we freeze?

How much heat can we get into the ice between -14 and 0 ?
heat into ice to raise to the melting point
= .035 * 2090 * 14 = 1024 Joules
so how much water at 0 deg can we freeze
with that .1024 * 10^4 Joules
we need 33.5 *10^4 Joules/kg
so we can freeze
.1024/33.5 = .003 kg or 3 grams
so we end up with
35+3 = 38 grams of ice.

Well, well, well! We have an icy situation here, don't we? Let's break it down and find out how much ice is left in the calorimeter!

First things first, we need to calculate the heat lost by the ice as it warms up from -14°C to 0°C. The specific heat capacity of ice is given as 2090 J/(kg K), and the mass of the ice is 35 g (or 0.035 kg). So, the heat lost by the ice can be calculated as:

Q1 = specific heat capacity × mass × change in temperature
= 2090 J/(kg K) × 0.035 kg × (0°C - (-14°C))

Now, let's calculate the heat lost by the water as it cools down from 0°C to the same final temperature as the ice. The specific heat capacity of water is approximately 4186 J/(kg K), and the mass of the water is given as 400 g (or 0.4 kg). Thus, the heat lost by the water can be calculated as:

Q2 = specific heat capacity × mass × change in temperature
= 4186 J/(kg K) × 0.4 kg × (0°C - (-14°C))

Now comes the interesting part. When all the ice has melted and the system reaches equilibrium, the heat gained by the resulting water is equal to the heat lost by the ice and the water. So, we need to calculate the heat gained by the water as it warms up from 0°C to the final temperature.

Here's where the latent heat of fusion of water comes into play! The latent heat of fusion of water is 33.5 x 10^4 J/kg. We don't know the mass of the water that freezes, but we can calculate it using the principle of energy conservation. The heat gained by the water can be calculated as:

Q3 = mass of water that freezes × latent heat of fusion of water
= mass of water that freezes × (33.5 x 10^4 J/kg)

According to the principle of energy conservation, we can then say that:

Q1 + Q2 = Q3

Now, using these equations and some algebra, we can solve for the mass of the water that freezes (or, in other words, the remaining ice in the calorimeter). So, drumroll, please...

The mass of ice left in the calorimeter is approximately 14.4 g. Voila!

To determine how much ice is left in the calorimeter at equilibrium, we need to calculate the heat transferred between the ice and water until they reach the same final temperature. We can break this process down into two steps: heating the ice to 0°C, and then melting the ice.

Step 1: Heating the ice to 0°C
First, we need to calculate the heat required to raise the temperature of the ice from -14°C to 0°C. We can use the formula:

Q1 = m1 * c1 * ΔT1

Where:
- Q1 is the heat transferred
- m1 is the mass of ice
- c1 is the specific heat capacity of ice
- ΔT1 is the change in temperature

In this case:
- m1 = 35 g (0.035 kg)
- c1 = 2090 J/(kg K)
- ΔT1 = 0°C - (-14°C) = 14°C

Substituting these values into the formula:

Q1 = (0.035 kg) * (2090 J/(kg K)) * (14 K) = 10085 J

Step 2: Melting the ice
Next, we need to calculate the heat required to melt the ice at its melting point of 0°C. The formula for this phase change is:

Q2 = m2 * Lf

Where:
- Q2 is the heat transferred
- m2 is the mass of ice (remaining after heating to 0°C)
- Lf is the latent heat of fusion of water

In this case:
- m2 is the remaining mass of ice, which we need to find
- Lf = 33.5 x 10^4 J/kg

Substituting these values into the formula, we get:

Q2 = m2 * (33.5 x 10^4 J/kg)

The total heat transferred during the process is equal to the sum of Q1 and Q2. Since it is a closed system, the heat gained by the water is equal to the heat lost by the ice:

Q1 + Q2 = m3 * c3 * ΔT3

Where:
- m3 is the mass of water
- c3 is the specific heat capacity of water
- ΔT3 is the change in temperature (final temperature - initial temperature)

In this case:
- m3 = 400 g (0.4 kg)
- c3 = 4186 J/(kg K) (specific heat capacity of water)
- ΔT3 = 0°C - (-14°C) = 14°C

Substituting these values into the formula:

10085 J + m2 * (33.5 x 10^4 J/kg) = (0.4 kg) * (4186 J/(kg K)) * (14 K)

Solving for m2 (mass of remaining ice), we can rearrange the equation:

m2 = [ (0.4 kg) * (4186 J/(kg K)) * (14 K) - 10085 J ] / (33.5 x 10^4 J/kg)

Calculating this expression will give us the mass of the remaining ice.

To solve this problem, we need to consider the energy exchange that takes place during the process.

First, let's calculate the heat required to raise the temperature of the ice to 0°C:

Q1 = mass of ice × specific heat of ice × change in temperature
Q1 = 35 g × 2090 J/(kg·K) × (0°C - (-14°C))
Q1 = 35 g × 2090 J/(kg·K) × 14°C
Q1 = 35 g × 2090 J/(kg·K) × 14 K (note that 1°C = 1 K)
Q1 = 35 g × 2090 J/K × 14

Next, let's calculate the heat required to melt the ice completely:

Q2 = mass of ice × latent heat of fusion of water
Q2 = 35 g × (33.5 × 10^4 J/kg)

Now, let's calculate the heat released by the water while it cools down from 0°C to the final equilibrium temperature:

Q3 = mass of water × specific heat of water × change in temperature
Q3 = 400 g × 4186 J/(kg·K) × 0°C

In equilibrium, the heat gained by the water (Q3) is equal to the heat lost by the ice (Q1 + Q2). Therefore:

Q3 = Q1 + Q2
400 g × 4186 J/(kg·K) × 0°C = 35 g × 2090 J/K × 14 K + 35 g × (33.5 × 10^4 J/kg)

Now, let's solve for the mass of ice that remains:

35 g × (33.5 × 10^4 J/kg) = 400 g × 4186 J/(kg·K) × 0°C - 35 g × 2090 J/K × 14 K

Simplifying the equation, we can find the final mass of the ice:

35 g × (33.5 × 10^4 J/kg) = (400 g × 4186 J/(kg·K) × 0°C) - (35 g × 2090 J/K × 14 K)

35 g × (33.5 × 10^4 J/kg) = 0 J + 35 g × 2090 J/K × 14 K

35 g × (33.5 × 10^4 J/kg) = 35 g × 2090 J/K × 14 K

(33.5 × 10^4 J/kg) = 2090 J/K × 14 K

33.5 × 10^4 J/kg = 2090 J/K × 14 K

33.5 × 10^4 J/kg = 29260 J

Now, let's solve for the mass of ice:

mass of ice = (33.5 × 10^4 J/kg) / (2090 J/kg·K)
mass of ice ≈ 160.77 g

Therefore, approximately 160.77 g of ice will remain in the calorimeter at equilibrium.