
The initial and final temperatures of water as recorded by an observer are $ {\left( {4.06 \pm 0.2} \right)^ \circ }C $ and $ {\left( {78.9 \pm 0.3} \right)^ \circ }C $ . Calculate the rise in temperature with proper error.
Answer
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Hint :In order to solve this question, we are going to subtract the temperature $ {\theta _1} $ from $ {\theta _2} $ , then the resulting error difference is also calculated by adding $ \Delta {\theta _1} $ and $ \Delta {\theta _2} $ , thus, putting a sign of plus minus for the error then the difference plus minus the error of the rise of temperature gives us the solution.
The rise of the temperature is given by the difference of the two temperatures as
$ \theta = {\theta _2} - {\theta _1} $
And that for the error in the temperatures measurement is given by the formula
$ \Delta \theta = \pm \left( {\Delta {\theta _1} + \Delta {\theta _2}} \right) $
Complete Step By Step Answer:
Let us start solving this problem by taking the initial and the final temperature values, we are given that the initial and the final temperatures of water recorded by the observer are as follows:
$ {\theta _1} = {\left( {4.06 \pm 0.2} \right)^ \circ }C \\
{\theta _2} = {\left( {78.9 \pm 0.3} \right)^ \circ }C \\ $
The rise in temperature is calculated by subtracting $ {\theta _1} $ from $ {\theta _2} $ ,
$ \theta = {\theta _2} - {\theta _1} \\
\Rightarrow \theta = 78.9 - 4.06 = {74.84^ \circ }C \\ $
The error in the rise in the temperature is calculated by adding $ \Delta {\theta _1} $ and $ \Delta {\theta _2} $
$ \Delta \theta = \pm \left( {\Delta {\theta _1} + \Delta {\theta _2}} \right) \\
\Rightarrow \Delta \theta = \pm \left( {0.2 + 0.3} \right) \\
\Rightarrow \Delta \theta = \pm 0.5 \\ $
Thus, the proper rise in temperature along with the proper error is given by
$ {\left( {74.84 \pm 0.5} \right)^ \circ }C $
Note :
The observer records the temperature with precision, however there are always some chances of error in measurement even with automatic machines, then the observer is bound to cause the error. If there is an error in the measurement of temperature, then its rise has to have some of the error. All types of errors are always additive in nature, for rise for fall, the errors are always added.
The rise of the temperature is given by the difference of the two temperatures as
$ \theta = {\theta _2} - {\theta _1} $
And that for the error in the temperatures measurement is given by the formula
$ \Delta \theta = \pm \left( {\Delta {\theta _1} + \Delta {\theta _2}} \right) $
Complete Step By Step Answer:
Let us start solving this problem by taking the initial and the final temperature values, we are given that the initial and the final temperatures of water recorded by the observer are as follows:
$ {\theta _1} = {\left( {4.06 \pm 0.2} \right)^ \circ }C \\
{\theta _2} = {\left( {78.9 \pm 0.3} \right)^ \circ }C \\ $
The rise in temperature is calculated by subtracting $ {\theta _1} $ from $ {\theta _2} $ ,
$ \theta = {\theta _2} - {\theta _1} \\
\Rightarrow \theta = 78.9 - 4.06 = {74.84^ \circ }C \\ $
The error in the rise in the temperature is calculated by adding $ \Delta {\theta _1} $ and $ \Delta {\theta _2} $
$ \Delta \theta = \pm \left( {\Delta {\theta _1} + \Delta {\theta _2}} \right) \\
\Rightarrow \Delta \theta = \pm \left( {0.2 + 0.3} \right) \\
\Rightarrow \Delta \theta = \pm 0.5 \\ $
Thus, the proper rise in temperature along with the proper error is given by
$ {\left( {74.84 \pm 0.5} \right)^ \circ }C $
Note :
The observer records the temperature with precision, however there are always some chances of error in measurement even with automatic machines, then the observer is bound to cause the error. If there is an error in the measurement of temperature, then its rise has to have some of the error. All types of errors are always additive in nature, for rise for fall, the errors are always added.
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