Chemistry, 25.09.2019 01:10 uwauwa1234

Aplate is removed from an oven at temperature of 50◦ and placed in a large room at 20◦ . after 5 minutes the temperature of the plate is 40◦ . how much longer will it take to reach 30◦ ?

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Answer from: joejoefofana

Explanation:

The given data is as follows.

Initial temperature ( $T_{1}$ ) = $50^{o}C$

Room temperature ( $T_{a}$ ) = $20^{o}C$

Time (t) = 5 min

Final temperature (T) = $40^{o}C$

According to the unsteady state equation,

$\frac{ln (T - T_{a})}{(T_{1} - T_{a})} = \frac{-h \times A \times t}{dV \times C}$

where, h = heat transfer coefficient

A = area

d = density

V = volume

C = specific heat capacity

All of these are constants and can be expressed as K

Therefore, K = $\frac{h \times A}{dV \times C}$

$\frac{ln (T - T_{a})}{(T_{1} - T_{a})} = -K \times t$

$\frac{ln (40 - 20)}{(50 - 20)} = -K \times 5$ 0.4055 = 5K

K = 0.0811

After 5 minutes, the new temperature will be as follows.

New temperature ( $T_{2}$ ) = $30^{o}C$

Time (t) = ?

Again, from the unsteady state conduction,

$\frac{ln (T_{2} - T_{a})}{(T_{i} - T_{a})} = [tex]\frac{-h \times A \times t}{dV \times C}$ = $-K \times t$

$\frac{ln [(30 - 20)}{(50 - 20)} = - 0.0811 \times t$

1.0986 = $0.0811 \times t$

t = 13.55 min

Thus, we can conclude that after 13.55 minutes the temperature reaches to $30^{o}C$ .

Answer from: Quest

the answer would be a

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

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