Problem 2.
If the temperature of the bread is 120oC, of the air is 30oC and K1=0.0366, K2=−0.0002 determine the temperature of the bread and air 60 minutes later.
(By linear algebraic method.)
We use the Newton's law of cooling,
M′(t)=K2(M(t)−T(t)),T′(t)=K1(M(t)−T(t)).
where T(t) is the temperature of the bread, M(t) is the temperature of the air and K1>0, K2<0 are constants. Denote T0:=T(0), M0:=M(0), M0<T0.
Introduce the vector x(t):=[M(t)T(t)], and the matrix A:=[K2−K2K1−K1]. Then we can write x′(t)=Ax(t). The general solution of this equation is x(t)=eAtc, where c∈R2. The eigenvalues and corresponding eigenvectors of A are λ1=−K1+K2,s1=[K2K11],λ2=0,s2=[11]. Since λ1≠λ2 are real numbers, x(t)=[s1s2][e−(K1−K2)t00e0t][c1c2]=[K2e−(K1−K2)tc1K1+c2e−(K1−K2)tc1+c2]. Using the initial conditions we have [M0T0]=x(0)=[K2c1K1+c2c1+c2]. Solving this system we obtain c1=−K1(M0−T0)K1−K2,c2=K1M0−K2T0K1−K2. Substituting back x(t)=[−K2e−(K1−K2)t(M0−T0)+K1M0−K2T0K1−K2−K1e−(K1−K2)t(M0−T0)+K1M0−K2T0K1−K2]. We note that limt→∞x(t)=[K1M0−K2T0K1−K2K1M0−K2T0K1−K2]. For the given data we obtain x(t)=[30.489−0.489e−0.0368t89.511e−0.0368t+30.489] and x(60)=[30.43540.328].◼
Introduce the vector x(t):=[M(t)T(t)], and the matrix A:=[K2−K2K1−K1]. Then we can write x′(t)=Ax(t). The general solution of this equation is x(t)=eAtc, where c∈R2. The eigenvalues and corresponding eigenvectors of A are λ1=−K1+K2,s1=[K2K11],λ2=0,s2=[11]. Since λ1≠λ2 are real numbers, x(t)=[s1s2][e−(K1−K2)t00e0t][c1c2]=[K2e−(K1−K2)tc1K1+c2e−(K1−K2)tc1+c2]. Using the initial conditions we have [M0T0]=x(0)=[K2c1K1+c2c1+c2]. Solving this system we obtain c1=−K1(M0−T0)K1−K2,c2=K1M0−K2T0K1−K2. Substituting back x(t)=[−K2e−(K1−K2)t(M0−T0)+K1M0−K2T0K1−K2−K1e−(K1−K2)t(M0−T0)+K1M0−K2T0K1−K2]. We note that limt→∞x(t)=[K1M0−K2T0K1−K2K1M0−K2T0K1−K2]. For the given data we obtain x(t)=[30.489−0.489e−0.0368t89.511e−0.0368t+30.489] and x(60)=[30.43540.328].◼
(By Laplace transformation.)
We use the Newton's law of cooling,
M′(t)=K2(M(t)−T(t)),T′(t)=K1(M(t)−T(t)).
where T(t) is the temperature of the bread, M(t) is the temperature of the air and K1>0, K2<0 are constants. Denote T0:=T(0), M0:=M(0), M0<T0.
Introduce the vector x(t):=[M(t)T(t)], and the matrix A:=[K2−K2K1−K1]. Then we can write x′(t)=Ax(t). Let ˆx(s) denote the Laplace transform of x(t). Then taking the Laplace transform of the system we get sˆx(s)−x0=Aˆx(s), or equivalently (sE−A)ˆx(s)=x0, where E is the unit matrix. Since (sE−A)−1=[s+K1s(s+K1−K2)−K2s(s+K1−K2)K1s(s+K1−K2)s−K2s(s+K1−K2)], we obtain ˆx(s)=(sE−A)−1x0=[(s+K1)M0−K2T0s(s+K1−K2)K1M0+(s−K2)T0s(s+K1−K2)]=[K1M0−K2T0s(K1−K2)−K2(M0−T0)(s+K1−K2)(K1−K2)K1M0−K2T0s(K1−K2)−K1(M0−T0)(s+K1−K2)(K1−K2)].
Introduce the vector x(t):=[M(t)T(t)], and the matrix A:=[K2−K2K1−K1]. Then we can write x′(t)=Ax(t). Let ˆx(s) denote the Laplace transform of x(t). Then taking the Laplace transform of the system we get sˆx(s)−x0=Aˆx(s), or equivalently (sE−A)ˆx(s)=x0, where E is the unit matrix. Since (sE−A)−1=[s+K1s(s+K1−K2)−K2s(s+K1−K2)K1s(s+K1−K2)s−K2s(s+K1−K2)], we obtain ˆx(s)=(sE−A)−1x0=[(s+K1)M0−K2T0s(s+K1−K2)K1M0+(s−K2)T0s(s+K1−K2)]=[K1M0−K2T0s(K1−K2)−K2(M0−T0)(s+K1−K2)(K1−K2)K1M0−K2T0s(K1−K2)−K1(M0−T0)(s+K1−K2)(K1−K2)].
Applying the inverse Laplace transformation
we obtain
x(t)=[−K2e−(K1−K2)t(M0−T0)+K1M0−K2T0K1−K2−K1e−(K1−K2)t(M0−T0)+K1M0−K2T0K1−K2].
We note that
limt→∞x(t)=[K1M0−K2T0K1−K2K1M0−K2T0K1−K2].
For the given data we obtain
x(t)=[30.489−0.489e−0.0368t89.511e−0.0368t+30.489]
and
x(60)=[30.43540.328].◼
