An object is heated and then allowed to cool when the temperature is 70 deg. its rate of cooling is 3 deg, celcius perminute and when the temperature is 60 deg. celcius, the rate of cooling is 2.5 deg. celcius per minute. Determine the temperature of the surroundings
To determine the temperature of the surroundings, we can use the rate of cooling for the object and its change in temperature.
Let's assume the initial temperature of the object is \(T_0\) degrees Celsius and the temperature of the surroundings is \(T_s\) degrees Celsius.
Given:
Rate of cooling when temperature is 70°C = 3°C per minute
Rate of cooling when temperature is 60°C = 2.5°C per minute
To find the temperature of the surroundings (\(T_s\)), we will set up an equation using the two given rates of cooling.
Step 1: Determine the rate of change of temperature (\(\Delta T\)) between the two given temperatures (70°C and 60°C).
\(\Delta T = 70 - 60 = 10°C\)
Step 2: Determine the change in time (\(\Delta t\)) based on the given rates of cooling.
For the rate of cooling at 70°C:
Rate of cooling = \(\frac{{\Delta T}}{{\Delta t}}\)
3 = \(\frac{{10}}{{\Delta t_1}}\)
For the rate of cooling at 60°C:
Rate of cooling = \(\frac{{\Delta T}}{{\Delta t}}\)
2.5 = \(\frac{{10}}{{\Delta t_2}}\)
Step 3: Solve the two equations for \(\Delta t_1\) and \(\Delta t_2\).
\(\Delta t_1 = \frac{{10}}{{3}}\) minutes
\(\Delta t_2 = \frac{{10}}{{2.5}}\) minutes
Step 4: Determine the average rate of cooling (\(\bar{r_c}\)) during the temperature change from 70°C to 60°C.
\(\bar{r_c}\) = \(\frac{{2 \times r_1 \times \Delta t_1 + 2 \times r_2 \times \Delta t_2}}{{\Delta t_1 + \Delta t_2}}\)
Since the rate of cooling is constant, we can assume it is the same during the entire cooling process. Therefore, we can use the average rate of cooling to find the temperature of the surroundings.
\(\bar{r_c}\) = \(\frac{{2 \times 3 \times \frac{{10}}{{3}} + 2 \times 2.5 \times \frac{{10}}{{2.5}}}}{{\frac{{10}}{{3}} + \frac{{10}}{{2.5}}}}\)
Simplifying the equation:
\(\bar{r_c}\) = \(\frac{{20}}{{\frac{{60}}{{3}} + \frac{{60}}{{2.5}}}}\)
Step 5: Calculate the average rate of cooling (\(\bar{r_c}\)).
\(\bar{r_c}\) ≈ 2.727°C per minute
Since the surroundings absorb heat from the object, the rate of cooling is equal to the rate of heat transfer from the object to the surroundings. Thus, the temperature of the surroundings (\(T_s\)) can be calculated using the formula:
\(\bar{r_c} = \frac{{T_0 - T_s}}{{\Delta t}}\)
Rearranging the formula to solve for \(T_s\):
\(T_s = T_0 - \bar{r_c} \times \Delta t\)
Substituting the values:
\(T_s = 70 - 2.727 \times \frac{{10}}{{3}}\)
Evaluating the equation:
\(T_s \approx 70 - 9.09\)
Therefore, the temperature of the surroundings is approximately 60.91°C.
To determine the temperature of the surroundings, we can use Newton's Law of Cooling, which states that the rate of heat loss (cooling) of an object is directly proportional to the difference in temperature between the object and its surroundings.
The formula for Newton's Law of Cooling is:
dT/dt = -k (T - Ts)
Where:
dT/dt is the rate of cooling of the object
T is the temperature of the object
Ts is the temperature of the surroundings
k is the cooling constant
We are given that the rate of cooling when the temperature is 70°C is 3°C per minute, and when the temperature is 60°C, the rate of cooling is 2.5°C per minute. We can now set up two equations using the given information.
Equation 1:
3 = -k (70 - Ts)
Equation 2:
2.5 = -k (60 - Ts)
We can now solve this system of equations to find the value of Ts, which represents the temperature of the surroundings.
First, rearrange Equation 1 and Equation 2 to isolate the term (Ts - 70) and (Ts - 60), respectively:
Equation 1:
-70k = 3 - 3k
70k - 3k = 3
67k = 3
k = 3/67
Equation 2:
-60k = 2.5 - 2.5k
60k - 2.5k = 2.5
57.5k = 2.5
k = 2.5/57.5
Now that we have the value of k in both equations, we can set them equal to each other and solve for Ts:
3/67 = 2.5/57.5
(3/67) * 57.5 = 2.5
172.5/67 = 2.5
172.5 = 2.5 * 67
172.5 = 167.5
Therefore, the temperature of the surroundings, Ts, is approximately 167.5 degrees Celsius.