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Question

Chemistry Question on Some basic concepts of chemistry

The temperature of 32C32^{\circ}C is equivalent to

A

89.6F89.6^{\circ}F

B

85.6F85.6^{\circ}F

C

70F70^{\circ}F

D

69F69^{\circ}F

Answer

89.6F89.6^{\circ}F

Explanation

Solution

The temperature of F{ }^{\circ} F and C{ }^{\circ} C are related to each other by the following relationship:

F=95(C)+32^{\circ} F =\frac{9}{5}\left({ }^{\circ} C \right)+32
F=9×325+32=57.6+32{ }^{\circ} F =\frac{9 \times 32}{5}+32= 57.6+32
F=89.6{ }^{\circ} F =89.6

The temperature of 32C32^{\circ} C is equivalent to 89.6F89.6^{\circ} F.