Answer :
To find the initial temperature of the solid object when it was dropped into the pond, we will follow a series of steps based on the given information.
### Given:
- Time, [tex]\( t = 4 \)[/tex] minutes
- Temperature after 4 minutes, [tex]\( f(4) = 35 \)[/tex] degrees Celsius
- Constant [tex]\( k = 0.0399 \)[/tex]
- Temperature of the pond [tex]\( = 20 \)[/tex] degrees Celsius
### Equation:
The temperature of the object as a function of time is given by:
[tex]\[ f(t) = C e^{-kt} + 20 \][/tex]
where [tex]\( C \)[/tex] is the constant we need to determine based on initial conditions.
### Step-by-Step Solution:
1. Substitute the given values into the equation to determine [tex]\( C \)[/tex]:
[tex]\[ f(4) = 35 \][/tex]
[tex]\[ 35 = C e^{-0.0399 \times 4} + 20 \][/tex]
2. Evaluate the exponent:
[tex]\[ 35 = C e^{-0.1596} + 20 \][/tex]
3. Isolate [tex]\( C \)[/tex]:
[tex]\[ 35 - 20 = C e^{-0.1596} \][/tex]
[tex]\[ 15 = C e^{-0.1596} \][/tex]
4. Solve for [tex]\( C \)[/tex]:
[tex]\[ C = \frac{15}{e^{-0.1596}} \][/tex]
5. Evaluate the denominator:
[tex]\[ C = 15 \times e^{0.1596} \][/tex]
6. Now, to find the initial temperature of the object, substitute [tex]\( t = 0 \)[/tex] into the original equation:
[tex]\[ f(0) = C e^{-0.0399 \times 0} + 20 \][/tex]
[tex]\[ f(0) = C \times 1 + 20 \][/tex]
[tex]\[ f(0) = C + 20 \][/tex]
7. With [tex]\( C \)[/tex] evaluated from previous steps:
[tex]\[ f(0) = 15 \times e^{0.1596} + 20 \][/tex]
Upon plugging in the values and solving, the initial temperature of the object is found to be:
[tex]\[ \boxed{37.6} \][/tex]
### Given:
- Time, [tex]\( t = 4 \)[/tex] minutes
- Temperature after 4 minutes, [tex]\( f(4) = 35 \)[/tex] degrees Celsius
- Constant [tex]\( k = 0.0399 \)[/tex]
- Temperature of the pond [tex]\( = 20 \)[/tex] degrees Celsius
### Equation:
The temperature of the object as a function of time is given by:
[tex]\[ f(t) = C e^{-kt} + 20 \][/tex]
where [tex]\( C \)[/tex] is the constant we need to determine based on initial conditions.
### Step-by-Step Solution:
1. Substitute the given values into the equation to determine [tex]\( C \)[/tex]:
[tex]\[ f(4) = 35 \][/tex]
[tex]\[ 35 = C e^{-0.0399 \times 4} + 20 \][/tex]
2. Evaluate the exponent:
[tex]\[ 35 = C e^{-0.1596} + 20 \][/tex]
3. Isolate [tex]\( C \)[/tex]:
[tex]\[ 35 - 20 = C e^{-0.1596} \][/tex]
[tex]\[ 15 = C e^{-0.1596} \][/tex]
4. Solve for [tex]\( C \)[/tex]:
[tex]\[ C = \frac{15}{e^{-0.1596}} \][/tex]
5. Evaluate the denominator:
[tex]\[ C = 15 \times e^{0.1596} \][/tex]
6. Now, to find the initial temperature of the object, substitute [tex]\( t = 0 \)[/tex] into the original equation:
[tex]\[ f(0) = C e^{-0.0399 \times 0} + 20 \][/tex]
[tex]\[ f(0) = C \times 1 + 20 \][/tex]
[tex]\[ f(0) = C + 20 \][/tex]
7. With [tex]\( C \)[/tex] evaluated from previous steps:
[tex]\[ f(0) = 15 \times e^{0.1596} + 20 \][/tex]
Upon plugging in the values and solving, the initial temperature of the object is found to be:
[tex]\[ \boxed{37.6} \][/tex]
