When handling chemicals and solvents, technicians are recommended to

Match each term in the second column with its correct definition in the first column.

Levart [38]

Answer:

The filled in the black answers is:

1. work book

2. Spreadsheet

3. Cell

4. Sheet tabs

5. Column

6. Row

7.Cell content

8. data

9. Formula

10. Constant value

11. Number value

12. Cell address

7 0

3 years ago

A 4 cm diameter sphere of copper is initially at a temperature of 95 °C. It is placed in a very large water bath at time t- 0. T

avanturin [10]

Answer:112.376 s

Explanation:

Given

$T_i=95^{\circ}C$

$T_f=35^{\circ}C$

$T_\infty \left ( ambient\right )=25^{\circ}C$

$h=400 watts/\left ( m^{2}^{\circ}C\right )$

$c=0.385 J/\left ( m^2^{\circ}C\right )$

$\rho =9 gm/cm^{3}$

Using Newton's law of cooling

$\frac{T_i-T_{\infty}}{T-T_{\infty}}$ = $e^{\frac{ht}{\rho L_{c}c}}$

$\frac{95-25}{35-25}$ = $e^{\frac{400\times 3\times 10^{-4}\times t}{9\times 2\times 0.385}}$

7= $e^{1.7316\times 10^{-2}\times t}$

Taking log both side

t=112.376sec

4 0

3 years ago

Carbon dioxide (CO2) is compressed in a piston-cylinder assembly from p1 = 0.7 bar, T1 = 320 K to p2 = 11 bar. The initial volum

tekilochka [14]

Answer:

$W_{12}=-53.9056KJ$

Part A:

$Q=-7.03734 KJ/Kg$ (-ve sign shows heat is getting out)

Part B:

$Q=1.5265KJ/Kg$ (Heat getting in)

The value of Q at constant specific heat is approximately 361% in difference with variable specific heat and at constant specific heat Q has opposite direction (going in) than Q which is calculated in Part B from table A-23. So taking constant specific heat is not a good idea and is questionable.

Explanation:

Assumptions:

Gas is ideal
System is closed system.
K.E and P.E is neglected
Process is polytropic

Since Process is polytropic so $W_{12} =\frac{P_{2}V_{2}-P_{1}V_{1}}{1-n}$

Where n=1.25

Since Process is polytropic :

$\frac{V_{2}}{V_{1}}=(\frac{P_{1}}{P_{2}})^{\frac{1}{1.25}} \\V_{2}= (\frac{P_{1}}{P_{2}})^{\frac{1}{1.25}} *V_{1}$

$V_{2}= (\frac{0.7}{11})^{\frac{1}{1.25}} *0.262\\V_{2}=0.028924 m^3$

Now, $W_{12} =\frac{P_{2}V_{2}-P_{1}V_{1}}{1-n}$

$W_{12} =\frac{11*0.028924-0.7*0.262}{1-1.25}(\frac{10^{5}N/m^2}{1 bar})(\frac{1 KJ}{10^{3}Nm})$

$W_{12}=-53.9056KJ$

We will now calculate mass (m) and Temperature T_2.

$m=\frac{P_{1}V_{1}}{RT_{1}}\\ m=\frac{0.7*0.262}{\frac{8.314KJ}{44.01Kg.K}*320}(\frac{10^{5}N/m^2}{1 bar})(\frac{1 KJ}{10^{3}Nm})\\m=0.30338Kg$

$T_{2} =\frac{P_{2}V_{2}}{Rm}\\ m=\frac{11*0.028924}{\frac{8.314KJ}{44.01Kg.K}*0.30338}(\frac{10^{5}N/m^2}{1 bar})(\frac{1 KJ}{10^{3}Nm})\\T_{2} =555.14K$

Part A:

According to energy balance::

$Q=mc_{v}(T_{2}-T_{1})+W_{12}$

From A-20, C_v for Carbon dioxide at 300 K is 0.657 KJ/Kg.k

$Q=0.30338*0.657(555.14-320)+(-53.9056)$

$Q=-7.03734 KJ/Kg$ (-ve sign shows heat is getting out)

Part B:

From Table A-23:

$u_{1} at 320K = 7526 KJ/Kg$

$u_{2} at 555.14K = 15567.292$ (By interpolation)

$Q=m(\frac{u(T_{2})-u(T_{1})}{M} )+W_{12}$

$Q=0.30338(\frac{15567.292-7526}{44.01} )+(-53.9056)$

$Q=1.5265KJ/Kg$ (Heat getting in)

The value of Q at constant specific heat is approximately 361% in difference with variable specific heat and at constant specific heat Q has opposite direction (going in) than Q which is calculated in Part B from table A-23. So taking constant specific heat is not a good idea and is questionable.

7 0

4 years ago

Consider a refrigerator that consumes 320 W of electric power when it is running. If the refrigerator runs only one quarter of t

Semenov [28]

Answer:

B. $5.18

Explanation:

Cost of electricity per kWh = $0.09

Power consumption of refrigerator = 320W = 320/1000 = 0.32kW

In a month (30 days) the refrigerator works 1/4 × 30 days = 7.5 days = 7.5 × 24 hours = 180 hours

Energy consumed in 180 hours = 0.32kW × 180h = 57.6kWh

Cost of electricity of 57.6kWh energy consumed by the refrigerator = 57.6 × $0.09 = $5.18

3 0

4 years ago

Read 2 more answers

Both model building codes and NFPA __________ can be used to determine the type of construction used in a building.

Alex777 [14]

Answer:

Both model building codes and NFPA 220 can be used to determine the type of construction used in a building.

8 0

3 years ago