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3 years ago

A 0.080 kg robin sits on a perch 6.0 meters above the ground. What is the potential energy of the robin

Chemistry

1 answer:

Nesterboy [21]3 years ago

6 0

Answer:

4.704J

Explanation:

The following data were obtained from the question:

m = 0.080kg

h = 6.0m

g = 9.8m/s^2

P.E =?

P.E = mgh

P.E = 0.08 x 9.8 x 6

P.E = 4.704J

Therefore, the potential energy of the robin is 4.704J

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A nuclear reactor core must stay at or below 95 °C to remain in good working condition. Cool water at a temperature of 10 °C is

aliina [53]

Answer:

$\large \boxed{\text{67 000 g}}$

Explanation:

This is a problem in calorimetry — the measurement of the quantities of heat that flow from one object to another.

It is based on the Law of Conservation of Energy — Energy can be transformed from one type to another, but it cannot be destroyed or created.

If heat flows out of the reactor (negative), the same amount of heat must flow into the water (positive).

Since there is no change in total energy,

heat₁ + heat₂ = 0

The symbol for the quantity of heat transferred is q, so we can rewrite the word equation as

q₁ + q₂ = 0

The formula for the heat absorbed or released by an object is

q = mCΔT, where

m = the mass of the sample

C = the specific heat capacity of the sample, and

ΔT = T_f - T_i = the change in temperature

1. Equation

There are two heat flows in this problem,

heat released by reactor + heat absorbed by water = 0

q₁ + q₂ = 0

q₁ + m₂C₂ΔT₂ = 0

2. Data:

q₁ = -23 746 kJ

m₂ = ?; C₂ = 4.184 J°C⁻¹g⁻¹; T_f = 95 °C; T_i = 10 °C

3. Calculations

(a) Convert kilojoules to joules

$q_{1} = -\text{23 746 kJ} \times \dfrac{\text{1000 J}}{\text{1 kJ}} = -\text{23 746 000 J}$

(b) ΔT

ΔT₂ = T_f - T_i = 95 °C - 10 °C = 85 °C

(c) m₂

$\begin{array}{rcl}q_{1} + q_{2} & = & 0\\\text{-23 746 000 J} + m_{2} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 85 \, ^{\circ}\text{C} & = & 0\\\text{-23 746 000 J} + 356m_{2} \text{J$\cdot$g}^{-1} & = & 0\\356m_{2} \text{g}^{-1} & = & 23746000\\m_2&=& \dfrac{23746000}{\text{356 g}^{-1}}\\\\ & = & \textbf{67000 g}\\\end{array}\\$

$\text{You must circulate $\large \boxed{\textbf{67 000 g}}$ of water each hour.}$

7 0

3 years ago

At 273 k and 1.00 x 10^-2 atm, the density of a gas is 1.24 x 10^-5 g/cm3.

bija089 [108]

Root mean square velocity is the square root of the mean of the squares of speeds of different molecules. From kinetic theory of gas, the formula of root mean square velocity=C $_{rms}$ = √ $\frac{3RT}{M}$ =√ $\frac{3PV}{M}$ =√ $\frac{3P}{d}$ , where, R= Universal gas constant, T= Absolute temperature, P= Pressure, V= Volume of gas, d= Density of gas.