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4 years ago

One gram of glass will rise 20° C when------------calories of heat are applied

Chemistry

1 answer:

astra-53 [7]4 years ago

6 0

Answer:

3.2 Calories

Explanation:

Here we will use a formula

Heat added in calories = Mass of glass x Increase in temperature x specific heat of glass

As we know that, specific heat is the amount of energy required to raise the temperature of one gram of any substance by 1°C. It has a constant value for every substance and for glass the specific heat is .16 calories/gm

Incorporating the values of mass (m), temperature(T) and specific heat (c) in formula.

calories (small calories) = l.0 g x 20 degrees x .16 calories/gm/°C

= 3.2 calories

Hope it helps:)

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What is the highest energy to which doubly ionized helium atoms (alpha particles) can be accelerated in a DC accelerator with 3

pentagon [3]

Answer:

Highest energy will be equal to $9.6\times 10^{-13}J$

Explanation:

Charged on doubly ionized helium atom $q=2e=2\times 1.6\times 10^{-19}C=3.2\times 10^{-19}C$

It is accelerated with maximum voltage of 3 MV

So voltage $V=3\times 10^6volt$

Now energy is given by $E=qV=3.2\times 10^{-19}\times 3\times 10^6=9.6\times 10^{-13}J$

So highest energy will be equal to $9.6\times 10^{-13}J$

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3 years ago

How many molecules are in 1.5 mole of glucose

GalinKa [24]

Answer:

In one mole of glucose, there are

6.022×1023

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Explanation:

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3 years ago

A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce

spin [16.1K]

Explanation:

The given data is as follows.

Volume of lake = $15 \times 10^{6} m^{3}$ = $15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}$

Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = $5.6 \times 15 \times 10^{9} mg$

= $84 \times 10^{9}$ mg

= $84 \times 10^{3}$ kg

Flow rate of river is 50 $m^{3} sec^{-1}$

Volume of water in 1 day = $50 \times 10^{3} \times 86400 liter$

= $432 \times 10^{7}$ liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are $2.9792 \times 10^{10} mg$ or $2.9792 \times 10^{4} kg$

Flow rate of sewage = $0.7 m^{3} sec^{-1}$

Volume of sewage water in 1 day = $6048 \times 10^{4}$ liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = $1.8144 \times 10^{10} mg$ or $1.8144 \times 10^{4}kg$

Therefore, total concentration of lake after 1 day = $\frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l$

= 6.8078 mg/l

$k_{D}$ = 0.2 per day

$L_{o}$ = 6.8078

Hence, $L_{liquid}$ = $L_{o}(1 - e^{-k_{D}t}$

$L_{liquid}$ = $6.8078 (1 - e^{-0.2 \times 1})$

= 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

= 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

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4 years ago

If faucet drips 5 mL of water each minute, what is the volume of water dripped at the end of five minutes?

kiruha [24]

If the faucet drops at 5 mL per minute you just have to do the following steps to find out.

5 mL for the dripping rate times 5 for the minutes.

The answer would result being 25mL

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3 years ago

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A kernel of popcorn contains a small amount of water and when heated the popcorn pops. How can you explain this to someone

Aliun [14]

When you put a popcorn kernel in a microwave, the microwave heats up the water. The water then evaporates, and the air wants to escape. There will be so much pressure that the skin/shell of the kernel will break, exposing the corn.

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3 years ago