Chapter 8: Problem 29

If \(10 \mathrm{~kg}\) of rock salt is placed in water, it dissolves at a rate proportional to the amount of salt still undissolved. If \(2 \mathrm{~kg}\) dissolve during the first 10 minutes, how long will it be until only 2 kg remain undissolved?

Short Answer

Expert verified

It will take approximately 72.62 minutes for only 2 kg to remain undissolved.

Step by step solution

Define the Problem Mathematically

Let the amount of salt undissolved at time t be denoted by \(y(t)\). The problem states that the rate of dissolution is proportional to the amount of salt undissolved. This can be represented by the differential equation: \[ \frac{dy}{dt} = -ky \] where \(k\) is the proportionality constant and \(y(t)\) is the amount of salt undissolved at time \(t\).

Solve the Differential Equation

To solve the differential equation, separate the variables: \[ \frac{1}{y}dy = -k dt \] Integrate both sides: \[ \int \frac{1}{y}dy = -\int k dt \] This simplifies to: \[ \ln|y| = -kt + C \] where \(C\) is the constant of integration. Solving for \(y\), we get: \[ y = e^{-kt+C} = Ce^{-kt} \] Thus, \[ y(t) = y_0 e^{-kt} \] where \(y_0\) is the initial amount of salt, which is 10 kg.

Find the Proportionality Constant k

We know that 2 kg of salt dissolves during the first 10 minutes, so when \(t = 10\), \(y(t) = 10 - 2 = 8\) kg. Plug this into the equation: \[ 8 = 10e^{-10k} \] Solving for k, we get: \[ e^{-10k} = \frac{8}{10} \] \[ -10k = \,\ln \left( \frac{8}{10} \right) \] \[ k = -\frac{\ln (0.8)}{10} \] \[ k \approx 0.0223 \]

Determine the Remaining Time for 2 kg Undissolved

We need to find the time \(t\) when \(y(t) = 2\) kg. Using the equation: \[ 2 = 10e^{-kt} \] Solving for \(t\), we get: \[ e^{-kt} = \frac{2}{10} \] \[ -kt = \,\ln (0.2) \] \[ t = -\frac{\ln(0.2)}{k} \] Substituting the value of \(k\): \[ t = -\frac{\ln(0.2)}{0.0223} \] \[ t \approx 72.62 \] Therefore, it will take approximately 72.62 minutes for only 2 kg to remain undissolved.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate of Dissolution

In this problem, we are examining how quickly the salt dissolves in water. The **rate of dissolution** is described by how fast the salt particles separate and dissolve into the water.

This rate is proportional to the amount of salt still left. In the problem, this means that the more salt is present, the faster it will dissolve initially. Over time, as the amount of salt reduces, the rate at which it dissolves also decreases.

Hence, the relationship is given by a differential equation which expresses this changing rate mathematically.

Separation of Variables

To solve the differential equation, we use **separation of variables**. This technique involves rearranging the equation so that all terms involving one variable are on one side and all terms involving the other variable are on the other side.

The given differential equation is:
ewline \[ \frac{dy}{dt} = -ky \]
First, separate the variables:
ewline \[ \frac{1}{y} dy = -k dt \]
Next, integrate both sides to find the solution:
ewline \[ \int \frac{1}{y} dy = \int -k dt \]
This leads to an expression involving logarithms.

Integration of Differential Equations

After separating the variables, we **integrate** each side to solve the differential equation. Integration helps find a function from its derivative.

Here, integrating both sides, we get:
ewline \[ \int \frac{1}{y} dy = -k t + C \]
where \(C\) is the constant of integration.
ewline Simplifying, we obtain:
ewline \[ \ln|y| = -kt + C \]
By exponents, this equation translates to:
ewline \[ y = Ce^{-kt} \]
This represents the undissolved salt at any time t.

Proportionality Constant

A key part of solving this problem is determining the **proportionality constant** \(k\). This constant relates the rate of dissolution to the amount of salt. We use information given to calculate it.

Given that 2 kg of salt dissolves in the first 10 minutes, when \(t = 10\):
ewline \[ y(10) = 10 e^{-10k} \]
We know \( y(10) = 8 \) kg, so: \[ 8 = 10 e^{-10k} \]
Solving for \(k\) gives: \[ k = -\frac{\text{ln}(0.8)}{10} \]
Once \(k\) is known, it helps determine how long the dissolution will take to reach any specified amount.