Question Number 209599 by peter frank last updated on 16/Jul/24
Answered by Spillover last updated on 16/Jul/24
Answered by Spillover last updated on 16/Jul/24
Answered by Spillover last updated on 16/Jul/24
Answered by mr W last updated on 16/Jul/24
$$\Delta{T}=−\frac{\mathrm{1}.\mathrm{30}−\mathrm{1}.\mathrm{26}}{\mathrm{1}.\mathrm{26}×\mathrm{53}×\mathrm{10}^{−\mathrm{5}} }=−\mathrm{60}\:{K}=−\mathrm{60}°{C} \\ $$$$\mathrm{20}°−\mathrm{60}°=−\mathrm{40}\:°{C} \\ $$$$\Rightarrow{at}\:−\mathrm{40}°{C}\:{sphere}\:{begins}\:{to}\:{float}. \\ $$
Commented by mr W last updated on 16/Jul/24
$${it}'{s}\:{assumed},\:{that}\:{the}\:{glass}\:{sphere}\: \\ $$$${keeps}\:{its}\:{temperature}\:{of}\:\mathrm{20}°{C}. \\ $$
Commented by peter frank last updated on 18/Jul/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{both} \\ $$