All categories

3 years ago

How many joules of heat are lost by 1000g of granite as it cools from 41.2 Celsius to -12.9 Celsius?

Chemistry

1 answer:

lilavasa [31]3 years ago

8 0

Answer: The amount of heat released is 42739 Joules

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

$Q=m\times c\times \Delta T$

Q = Heat released =?

c = heat capacity of granite = $0.790J/g^0C$

Initial temperature = $T_i$ = $41.2^0C$

Final temperature of the calorimeter = $T_f$ = $-12.9^0C$

Change in temperature , $\Delta T=T_f-T_i=(-12.9-41.2)^0C=-54.1^0C$

Putting in the values, we get:

$Q=1000g\times 0.790J/g^0C\times -54.1^0C=-42739J$

As heat comes out to be negative, that means the heat has been released and the amount of heat released is 42739 Joules

You might be interested in

At the beginning of the industrial revolution in 1850, the CO2concentration was 280 ppm. Today, it is 410 ppm.

Orlov [11]

Answer:

1. 2.04 W/m²

2. 1.63°C

Explanation:

The radiative force that the Earth receives comes from the Sun. When the Sun rays come to the surface, some of them are absorbed and then it is reflected in the space. The greenhouse gases (like CO2) blocks some of these rays, and then the surface stays warm. The excessive amount of these gases makes the surface warmer, which unbalance the climate on Earth.

1. The variation of the radiative forcing can be calculated based on the concentration of the CO2 by the equation:

ΔF = 5.35*ln(C/C0)

Where C is the final concentration, and C0 is the initial concentration.

ΔF = 5.35*ln(410/280)

ΔF = 2.04 W/m²

2. The temperature change in the Earth's surface caused by the variation of the radiative forcing can be calculated by:

ΔT = 0.8*ΔF

ΔT = 0.8*2.04

ΔT = 1.63 K = 1.63°C

6 0

3 years ago

When 0.100 mol of carbon is burned in a closed vessel with8.00

antoniya [11.8K]

Answer : The mass of carbon monoxide form can be 2.8 grams.

Solution : Given,

Moles of C = 0.100 mole

Mass of $O_2$ = 8.00 g

Molar mass of $O_2$ = 32 g/mole

Molar mass of CO = 28 g/mole

First we have to calculate the moles of $O_2$ .

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{8g}{32g/mole}=0.25moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2C+O_2\rightarrow 2CO$

From the balanced reaction we conclude that

As, 2 mole of $C$ react with 1 mole of $O_2$

So, 0.1 moles of $C$ react with $\frac{0.1}{2}=0.05$ moles of $O_2$

From this we conclude that, $O_2$ is an excess reagent because the given moles are greater than the required moles and $C$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CO$