Cooling Glass Of Water Lab Report
Using a cooling glass of water to determine the area and volume of the curve formed.
Research Question: To what extent can a cooling glass of water be modelled accurately and be used to find the area under the curve and volume formed by curve?
Data Collection: In order to produce an exponentially decaying graph, I had to first heat a glass of water for a certain period of time {150 ml of water}. After the water was heated a temperature probe, connected to logger pro and laptop, was placed inside the beaker and the change in temperature in contrast with time was noted down. The logger pro was set to collect data every 0.5 sec for 1300 sec in order to produce highly accurate and precise readings. It was noted that the temperature …show more content…
Microsoft excel was used in order to plot the graph, where x axis represents time and y axis represents the temperature. I know that the glass of water will never cool down beyond the room temperature and so we have an asymptote at any value just less that the room temperature. In this case, the room temperature while conducting the experiment as stable at 23.01°C. Thus I assumed an asymptote at 23°C.
1st equation:
An equation for an exponentially decaying graph is in the form of: y = a + b (e –m x)
Where: a= Vertical Translations (Ambient Temperature), b= Dilations, m= Gradient x= Time (t), y= Temperature (T (t))
From the data, we know that the value of ‘a’ is always equal to 23 °C. The original equation is equal to:
T (t) = 23 + b (e –m t) This part of the investigation had some issues. I had formed the equation but was confused on how to solve it. Later I plugged in two values to solve the equation. To make e –mt calculation easier and approachable, I converted the equation using natural …show more content…
From the table below, we can see that the mean difference between the actual and calculated data values is 1.205858454 which is higher than that of the 1st equation (0.1869). The table above represents the actually obtained data and the data obtained from the 2nd equation. The difference values help us understand the difference between two values and thus helps us predict the accuracy of the data obtained from the first equation. Mean difference is the average difference between theoretical and actual values. These values will later be compared with values from other equation to determine the equation closest to the actual data values.
3rd Equation: It was during one of my physics class, I came across the newton’s law of cooling which states that ‘The rate of change of temperature is directly proportional difference between the temperature of the body and the surroundings. Mathematically it meant that: However we need to solve the equation to suit our needs.
T’ (t) = –k (T (t) – a) (a = Ta = 23 = room temperature)
∫▒〖( (T' (t)) )/( (T (t) – 23) ) dt〗 =∫▒〖-k dt〗
Substitute u = T (t) – 23 and du = T′ (t) dt:
LHS
= ∫▒〖u^(-1) du〗
= ln |u| + C
= ln |T (t) − 23| +
Research Question: To what extent can a cooling glass of water be modelled accurately and be used to find the area under the curve and volume formed by curve?
Data Collection: In order to produce an exponentially decaying graph, I had to first heat a glass of water for a certain period of time {150 ml of water}. After the water was heated a temperature probe, connected to logger pro and laptop, was placed inside the beaker and the change in temperature in contrast with time was noted down. The logger pro was set to collect data every 0.5 sec for 1300 sec in order to produce highly accurate and precise readings. It was noted that the temperature …show more content…
Microsoft excel was used in order to plot the graph, where x axis represents time and y axis represents the temperature. I know that the glass of water will never cool down beyond the room temperature and so we have an asymptote at any value just less that the room temperature. In this case, the room temperature while conducting the experiment as stable at 23.01°C. Thus I assumed an asymptote at 23°C.
1st equation:
An equation for an exponentially decaying graph is in the form of: y = a + b (e –m x)
Where: a= Vertical Translations (Ambient Temperature), b= Dilations, m= Gradient x= Time (t), y= Temperature (T (t))
From the data, we know that the value of ‘a’ is always equal to 23 °C. The original equation is equal to:
T (t) = 23 + b (e –m t) This part of the investigation had some issues. I had formed the equation but was confused on how to solve it. Later I plugged in two values to solve the equation. To make e –mt calculation easier and approachable, I converted the equation using natural …show more content…
From the table below, we can see that the mean difference between the actual and calculated data values is 1.205858454 which is higher than that of the 1st equation (0.1869). The table above represents the actually obtained data and the data obtained from the 2nd equation. The difference values help us understand the difference between two values and thus helps us predict the accuracy of the data obtained from the first equation. Mean difference is the average difference between theoretical and actual values. These values will later be compared with values from other equation to determine the equation closest to the actual data values.
3rd Equation: It was during one of my physics class, I came across the newton’s law of cooling which states that ‘The rate of change of temperature is directly proportional difference between the temperature of the body and the surroundings. Mathematically it meant that: However we need to solve the equation to suit our needs.
T’ (t) = –k (T (t) – a) (a = Ta = 23 = room temperature)
∫▒〖( (T' (t)) )/( (T (t) – 23) ) dt〗 =∫▒〖-k dt〗
Substitute u = T (t) – 23 and du = T′ (t) dt:
LHS
= ∫▒〖u^(-1) du〗
= ln |u| + C
= ln |T (t) − 23| +