question_answer 28)

The temperature at 12 noon was \[\text{1}0{}^\circ \text{C}\] above zero. If it decreases at the rate of \[\text{2}{}^\circ \text{C}\] per hour until mid-night, at what time would the temperature be \[\text{8}{}^\circ \text{C}\] degrees below zero? What would temperature at mid-night?

Answer:

                Temperature at 12 noon =\[~+\text{1}0{}^\circ \text{C}\] Temperature at 1 P.M. = \[\text{1}0{}^\circ \text{C}-\text{2}{}^\circ \text{C}=\text{8}{}^\circ \text{C}\] Temperature at 2 P.M. = \[{{\text{8}}^{o}}\text{C}-\text{2}{{0}^{o}}\text{C}={{\text{6}}^{o}}\text{C}\] Temperature at 3 P.M. = \[{{\text{6}}^{o}}\text{C}-{{\text{2}}^{o}}\text{C}={{\text{4}}^{o}}\text{C}\] Temperature at 4 P.M. = \[{{\text{4}}^{o}}\text{C}-{{\text{2}}^{o}}\text{C}={{\text{2}}^{o}}\text{C}\] Temperature at 5 P.M. = \[{{\text{2}}^{o}}\text{C}-{{\text{2}}^{o}}\text{C}={{0}^{o}}\text{C}\] Temperature at 6 P.M. =  \[0{}^\circ \text{C}-\text{2}{}^\circ \text{C}=-\text{ 2}{}^\circ \text{C}\] Temperature at 7 P.M. = \[-\text{2}{}^\circ \text{C}-\text{2}{}^\circ \text{C}=-\text{ 4}{}^\circ \text{C}\] Temperature at 8 P.M. =\[-\text{ 4}{}^\circ \text{C}-\text{2}{}^\circ \text{C}=-\text{ 6}{}^\circ \text{C}\] Temperature at 9 P.M. =\[-\text{ 6}{}^\circ \text{C}-\text{2}{}^\circ \text{C}=-\text{ 8}{}^\circ \text{C}\] So, the temperature was \[\text{8}{}^\circ \text{C}\] below zero at 9 P.M. Temperature at 10 P.M. = \[-\text{ 8}{}^\circ \text{C}-\text{2}{}^\circ \text{C}\]\[=-\text{1}0{}^\circ \text{C}\] Temperature at 11 P.M. = \[-\text{ 1}0{}^\circ \text{C}-\text{2}{}^\circ \text{C}\]\[=-\text{12}{}^\circ \text{C}\] Temperature at 12 P.M. = \[-\text{12}{}^\circ \text{C}-\text{2}{}^\circ \text{C}\]\[=-\text{14}{}^\circ \text{C}\] So, the temperature at mid-night was 12 degrees below zero. Aliter: Difference between the temperatures \[\text{1}0{}^\circ \text{C}\] above zero and 8°C below zero \[=10{}^\circ \text{C}-(-\text{ 8}{}^\circ \text{C})=\text{1}0{}^\circ \text{C}+\text{8}{}^\circ \text{C}=\text{18}{}^\circ \text{C}\] Time for this degrees   \[=\text{18}\div \text{2}=\text{9}\] hours So, the temperature was \[\text{8}{}^\circ \text{C}\] below zero at 9 P.M. Difference between the times 12 noon and mid-night = 12 hours Degrees in temperature in 12 hours \[=\text{12}\times \text{2}\]\[=\text{24}{}^\circ \text{C}\] So, temperature at mid-night\[=\text{1}0{}^\circ \text{C}-\text{24}{}^\circ \text{C}\]\[~=-\text{14}{}^\circ \text{C}\].