# A 135 g sample of H20 at 85°C is cooled. The water loses a total of 15 kJ of energy in the coolingprocess. What is the final temperature of the water? The specific heat of water is 4.184 J/g.°C.A. 112°CB. 58°CC. 70°CD. 84°CE. 27°C

B. 58°C

Explanation:

Hello,

In this case, the relationship among heat, mass, specific heat and temperature for water is mathematically by:

In such a way, solving for the final temperature we obtain:

Therefore, we final temperature is computed as follows, considering that the involved heat is negative as it is lost for water:

Regards.

The final temperature of the water is approximately 58°C (Option B) after it has lost 15 kJ of energy in the cooling process.

### Explanation:

To calculate the final temperature of the water, we can use the formula of heat loss or gain, Q = mcΔT, where Q is the heat loss or gain, m is the mass, c is the specific heat, and ΔT is the change in temperature.

Setting the given values into our formula: -15,000 J = (135 g) * (4.184 J/g°C) * (T_final - 85°C).

Solving this equation for T_final gives us T_final ≈ 58°C. Hence, the suitable answer would be option B: 58°C.

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## Related Questions

What is the only gaseous component of the atmosphere whose concentration is routinely contained in all weather reports?

In weather report generally the temperature conditions and the extent of humidity is described. Humidity is nothing but presence of water in its vapour form. Hence the only gas which is being reported in a weather report must be water vapour. The other gases like oxygen carbon dioxide or noble gases or nitrogen are never mentioned in a normal weather report issued in public interest

A rock has a mass of 15 grams and a volume of 5 cm3 what is the density of the object￼￼

## 3.0 g/cm³

Explanation:

To find the density of a substance when given the mass and volume we use the formula

From the question

mass = 15 g

volume of rock = 5 cm³

The density of the substance is

We have the final answer as

### 3.0 g/cm³

Hope this helps you

The density of an object is calculated by dividing its mass by its volume. In this case, the rock's density is 3 g/cm3.

### Explanation:

The process to determine the density of an object involves dividing its mass by its volume. Here, the rock has a mass of 15 grams and a volume of 5 cm3. Thus, the density can be calculated by the formula:

Density = Mass / Volume.

Plugging the given numbers into this formula results in:

Density = 15 grams / 5 cm3.

Therefore, the density of the rock is 3 g/cm3.

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A car averages 27.5 miles per gallon

How much is each gallon or how far are you going is the question you should be asking

Abigail obtained 36.6 grams of calcium carbonate after performing a reaction. From her calculations, she knew she should have obtained 44.1 grams. What was her percent yield

it would be 1

Explanation:

1. What structural property of sodium 4-amino-1-naphthalenesulfonate makes it very soluble in water? 2. You will have to look up the structure of this compound and comment on why it is water-soluble. Simply stating that it's polar

1. Sodium 4-amino-1-naphthalenesulfonate makes it very soluble in water as it contains a hydrate salt sodium sulfate .

2. In the structure of this compound, sodium sulphate is polar in nature.

### Molecular formula:

The molecular structure of sodium 4-amino-1-naphthalenesulfonate is .

The polar part of the structure sodium sulfate makes sodium 4-amino-1-naphthalenesulfonate a hydrate salt. Salt are polar and are usually soluble in water.

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Answer: it contains a hydrate salt sodium sulfate NaO4S.

4-amino-1-naphthalenesulfonate is a sodium salt. Sodium sulfate is Polar.

Explanation:

The molecular structure of sodium 4-amino-1-naphthalenesulfonate is

C10H10NNaO4S

The polar part of the structure sodium sulfate NaO4S makes sodium 4-amino-1-naphthalenesulfonate a hydrate salt. Salt are polar and are usually soluble in water.