Gujarati
Hindi
10-1.Thermometry, Thermal Expansion and Calorimetry
normal

Maximum density of $H_2O$ is at temperature

A

$32\,^oF$

B

$4\,^oK$

C

$4\,^oF$

D

$39.2\,^oF$

Solution

Maximum density at $4^{\circ} \mathrm{C}$

$\frac{\mathrm{T}_{\mathrm{C}}-0}{100-0}=\frac{\mathrm{T}_{\mathrm{F}}-32}{212-32}$

$\frac{4}{100}=\frac{T_{F}-32}{180}$

$\boxed{{{\text{T}}_{\text{F}}} = {{39.2}^\circ }{\text{F}}}$

Standard 11
Physics

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