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MaRussiya [10]
3 years ago
9

Culver Corporation’s adjusted trial balance contained the following asset accounts at December 31, 2017: Cash $8,220, Land $40,8

00, Patents $16,000, Accounts Receivable $97,530, Prepaid Insurance $6,040, Inventory $34,900, Allowance for Doubtful Accounts $4,520, and Equity Investments (to be sold in the next quarter) $13,510. Prepare the current assets section of the balance sheet. (List Current Assets in order of liquidity.)
Business
1 answer:
Dovator [93]3 years ago
8 0

Answer:

See explanation Section

Explanation:

             Culver Corporation

 Balance Sheet (Current Asset only)

        As at December 31, 2017

Particulars                         $                         $

Cash                                                        $8,220                

Accounts Receivable $97,530

Less: Allowance for

<u>Doubtful Accounts       (4,520)           </u>   $93,010

Prepaid Insurance                                   $6,040

Inventory                                                $34,900

<u>Equity Investments                                  $13,510</u>

Current Assets                                     $155,680

Note: As equity investment will be sold in the next year, it is shown as current assets. Land and patents are property, plant, and equipment.

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Suppose 20.0 g pieces of gold and iron, both initially at 100oC, are added to different containers of water, both initially at 2
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Answer:

  • The final temperature in the container with Gold is 27.49 ⁰C
  • The final temperature in the container with Iron is  33.01  ⁰C

Therefore, the highest final temperature is obtained in the container with Iron.

Explanation:

Q = mcΔT

Where;

Q is the quantity of heat gained or lost

m is the mass of the metals or water

c is the specific heat capacity

ΔT is the change in temperature, T₂ - T₁

T₂  is the final temperature and T₁ is the initial temperature

Heat lost by metals at 100°C is equal to heat gained by water at 25°C

-Q_{metal} = Q_{water}

-Q_{Au} = Q_{H_2O} \\-Q_{Fe}  = Q_{H_2O}

Specific heat capacity of water = 4.18 J/g°C

Specific heat capacity of gold = 0.129 J/g°C

Specific heat capacity of iron  = 0.45 J/g°C

⇒For Gold

-20*0.129*(T₂ - 100) = 18*4.18 (T₂ - 25)

-2.58T₂ +258 = 75.24T₂  - 1881

77.82T₂  = 2139

T₂  = 2139/77.82

T₂ = 27.49 ⁰C

⇒For Iron

-20*0.450*(T₂ - 100) = 18*4.18 (T₂  - 25)

-9T₂  +900 = 75.24T₂ - 1881

84.24T₂ = 2781

T₂  = 2781/84.24

T₂ = 33.01  ⁰C

Therefore, the highest final temperature is obtained in the container with Iron.

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