How much greater is the light-collecting area of a 6-meter telescope than a 3-meter telescope?.

Solution A has a speciﬁc heat of 2.0 J/g◦C. Solution B has a speciﬁc heat of 3.8 J/g◦C. If equal masses of both solutions start

fgiga [73]

Answer: 2. Solution A attains a higher temperature.

Explanation: Specific heat simply means, that amount of heat which is when supplied to a unit mass of a substance will raise its temperature by 1°C.

In the given situation we have equal masses of two solutions A & B, out of which A has lower specific heat which means that a unit mass of solution A requires lesser energy to raise its temperature by 1°C than the solution B.

Since, the masses of both the solutions are same and equal heat is supplied to both, the proportional condition will follow.

We have a formula for such condition,

$Q=m.c.\Delta T$ .....................................(1)

where:

$\Delta T$ = temperature difference

Q= heat energy

m= mass of the body

c= specific heat of the body

Proving mathematically:

According to the given conditions

we have equal masses of two solutions A & B, i.e. $m_A=m_B$

equal heat is supplied to both the solutions, i.e. $Q_A=Q_B$

specific heat of solution A, $c_{A}=2.0 J.g^{-1} .\degree C^{-1}$

specific heat of solution B, $c_{B}=3.8 J.g^{-1} .\degree C^{-1}$

$\Delta T_A$ & $\Delta T_B$ are the change in temperatures of the respective solutions.

Now, putting the above values

$Q_A=Q_B$

$m_A.c_A. \Delta T_A=m_B.c_B . \Delta T_B\\\\2.0\times \Delta T_A=3.8 \times \Delta T_B\\\\ \Delta T_A=\frac{3.8}{2.0}\times \Delta T_B\\\\\\\frac{\Delta T_{A}}{\Delta T_{B}} = \frac{3.8}{2.0}>1$

Which proves that solution A attains a higher temperature than solution B.

7 0

3 years ago

How can a person detect infrared rays without an instrument?

natita [175]

So we want to know how can we detect infrared rays without an instrument. Infrared rays or heat, are a part of electromagnetic spectrum. We have specialized nerve cells in our skin called thermoreceptors that can detect differences in temperature that are produced by infrared part of EM spectrum.

7 0

3 years ago

Read 2 more answers

Fill in the blanks for the following:

storchak [24]

Answer:

a. 4.21 moles

b. 478.6 m/s

c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank

Explanation:

Volume of container = 100.0 L

Temperature = 293 K

pressure = 1 atm = 1.01325 bar

number of moles n = ?

using the gas equation PV = nRT

n = PV/RT

R = 0.08206 L-atm- $mol^{-1}$ $K^{-1}$

Therefore,

n = (1.01325 x 100)/(0.08206 x 293)

n = 101.325/24.04 = 4.21 moles

The equation for root mean square velocity is

Vrms = $\sqrt{\frac{3RT}{M} }$

R = 8.314 J/mol-K

where M is the molar mass of oxygen gas = 31.9 g/mol = 0.0319 kg/mol

Vrms = $\sqrt{\frac{3*8.314*293}{0.0319} }$ = 478.6 m/s

For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship

$\frac{Voxy}{Vnit}$ = $\sqrt{\frac{Mnit}{Moxy} }$

where

Voxy = root mean square velocity of oxygen = 478.6 m/s

Vnit = root mean square velocity of nitrogen = ?

Moxy = Molar mass of oxygen = 31.9 g/mol

Mnit = Molar mass of nitrogen = 14.00 g/mol

$\frac{478.6}{Vnit}$ = $\sqrt{\frac{14.0}{31.9} }$

$\frac{478.6}{Vnit}$ = 0.66

Vnit = 0.66 x 478.6 = 315.876 m/s

the root mean square velocity of the oxygen gas is

478.6/315.876 = 1.5 times the root mean square velocity of the nitrogen gas outside the tank

6 0

3 years ago

A heat engine running backward is called a refrigerator if its purpose is to extract heat from a cold reservoir. The same engine

Goryan [66]

Answer:

a) 2.85 kW

b) $ 432

c) $ 76.95

Explanation:

Average price of electricity = 1 $/40 MJ

Q = 20 kW

Heat energy production = 20.0 KJ/s

Coefficient of performance, K = 7

also

K=(QH)/Win

Now,

Coefficient of Performance, K = (QH)/Win = (QH)/P(in) = 20/P(in) = 7

where

P(in) is the input power

Thus,

P(in) = 20/7 = 2.85 kW

b) Cost = Energy consumed × charges

Cost = ($1/40000kWh) × (16kW × 300 × 3600s)

cost = $ 432

c) cost = (1$/40000kWh) × (2.85 kW × 200 × 3600s) = $76.95

4 0

3 years ago

calculate the efficiency of a light bulb that had an input of 400 j transfers 100j as a light and 300j as heat.

beks73 [17]

Efficiency = useful energy out / total energy in x 100
= 100/400 x 100
=0.25 x 100
= 25%

25%

6 0

3 years ago