## newton law of cooling

The rate at which a hot body loses heat is directly proportional to the difference between the temperature of the hot body and that of its surroundings and depends on the nature of the material and surface area of the body. Suppose m is the mass of a hot body and s is the specific heat at its initial temperature Q higher than its surroundings temperature Qo , dQ/dt is the rate of loss of heat dQ is the amount of heat lost by the hot body to its surroundings in a small interval of time.

1. Develop the differential equation of the hot body losing heat.

2. Enter the differential equation in MATLAB.

3. Solve the differential equation using MATLAB codes.

4. Plot the graph of log_10(Q – Qo) versus t using MATLAB. Then, interpret the results from the drawn graph.

5. Plot the graph of temperature (Q – Qo) versus time t using MATLAB. Interpret the results from the drawn figure.

6. Extract the MATLAB codes and figures for grading.The rate at which a hot body loses heat is directly proportional to the difference between the temperature of the hot body and that of its surroundings and depends on the nature of the material and surface area of the body. Suppose m is the mass of a hot body and s is the specific heat at its initial temperature Q higher than its surroundings temperature Qo , dQ/dt is the rate of loss of heat dQ is the amount of heat lost by the hot body to its surroundings in a small interval of time.

1. Develop the differential equation of the hot body losing heat.

2. Enter the differential equation in MATLAB.

3. Solve the differential equation using MATLAB codes.

4. Plot the graph of log_10(Q – Qo) versus t using MATLAB. Then, interpret the results from the drawn graph.

5. Plot the graph of temperature (Q – Qo) versus time t using MATLAB. Interpret the results from the drawn figure.

6. Extract the MATLAB codes and figures for grading. The rate at which a hot body loses heat is directly proportional to the difference between the temperature of the hot body and that of its surroundings and depends on the nature of the material and surface area of the body. Suppose m is the mass of a hot body and s is the specific heat at its initial temperature Q higher than its surroundings temperature Qo , dQ/dt is the rate of loss of heat dQ is the amount of heat lost by the hot body to its surroundings in a small interval of time.

1. Develop the differential equation of the hot body losing heat.

2. Enter the differential equation in MATLAB.

3. Solve the differential equation using MATLAB codes.

4. Plot the graph of log_10(Q – Qo) versus t using MATLAB. Then, interpret the results from the drawn graph.

5. Plot the graph of temperature (Q – Qo) versus time t using MATLAB. Interpret the results from the drawn figure.

6. Extract the MATLAB codes and figures for grading. differential equation MATLAB Answers — New Questions