How many molecules of h20 contain 2.0 x 10^22 molecules of water?

Suppose a substance has a heat of fusion equal to 45 calg and a specific heat of 0.75

Varvara68 [4.7K]

Answer:

The substance will be in liquid state at a temperature of 97.3 °C

Note: The question is incomplete. The complete question is given below :

Suppose a substance has a heat of fusion equal to 45 cal/g and a specific heat of 0.75 cal/g°C in the liquid state. If 5.0 kcal of heat are applied to a 50 g sample of the substance at a temperature of 24°C, what will its new temperate be? What state will the sample be in? (melting point of the substance = 27°C; specific heat of the solid =0.48 cal/g°C; boiling point of the substance = 700°C)

Explanation:

1.a) Heat energy required to raise the temperature of the substance to its melting point, H = mcΔT

Mass of solid sample = 50 g; specific heat of solid = 0.75 cal/g; ΔT = 27 - 24 = 3 °C

H = 50 × 0.75 × 3 = 112.5 calories

b) Heat energy required to convert the solid to liquid at its melting point at 27°C, H = m×l, where l = 45 cal/g

H = 50 × 45 = 2250 cal

c) Total energy used so far = 112.5 cal + 2250 cal = 2362.5 calories.

Amount of energy left = 5000 - 2362.5 = 2637.5 cal

The remaining energy is used to heat the liquid

H = mcΔT

Where specific heat of the liquid, c = 0.75 cal/g/°C, H = 2637.5 cal, ΔT = temperature change

2637.5 = 50 × 0.75 x ΔT

ΔT = 2637.5 / ( 50*0.75)

ΔT = 70.3 °C

Final temperature of sample = (70.3 + 27) °C = 97.3 °C

The substance will be in liquid state at a temperature of 97.3 °C

5 0

3 years ago

An aluminum can weighing 10 g absorbs 106.8 J of heat and warms by 12 degrees C. What is the specific heat of the aluminum can?

Sladkaya [172]

Answer:

$c_{Al} = 0.89\,\frac{J}{kg\cdot ^{\circ}C}$

Explanation:

The heating process is modelled after the First Law of Thermodynamics:

$Q = m \cdot c_{Al}\cdot \Delta T$

The specific heat of the aluminium can is:

$c_{Al} = \frac{Q}{m\cdot \Delta T}$

$c_{Al} = \frac{106.8\,J}{(10\,g)\cdot (12^{\circ}C)}$

$c_{Al} = 0.89\,\frac{J}{kg\cdot ^{\circ}C}$

8 0

3 years ago

What are the missing coefficients for the chemical equation <br> [ ] Ca + [ ] O2 → [ ] CaO

katen-ka-za [31]

Answer:

${ \sf{[ 2] Ca_{(s)} + [1 ] O_{2(g)} → [ 2] CaO _{(s)}}}$

4 0

3 years ago

Cl2 + AlBr3 → AlCl3 + Br2How many grams of Cl2 are needed to produce 19.03g of AlCl3?

VladimirAG [237]

Answer:

14.89 g

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

5 0

3 years ago

When the surface water temperature is still well below the water boiling temperature, some bubbles at the bottom tend to float u

irga5000 [103]

Answer:

The reasons why the seemingly floating bubbles disappear was that they tend to loss their latent heat to the water molecules at the surface water.

Explanation:

Heat energy has a considerable effect on the velocity of molecules including water. The water molecules below the container will receive much more heat energy than those above it. This heat energy in the form of specific heat capacity and latent heat that result in the increase in the speed of individual molecules of water and finally to the escape of the molecules to a colder region of the container, in this case the upper region. At the collision of the bottom water to the surface water, they tend to exchange their heat content, the hotter molecules will lose their heat to the cold ones. When the formerly hot molecules encounter this, it will result in lowering the temperature and consequentially to the reduction of their movement, once in the form of bubble, now become ordinary water. This convectional transfer of heat energy will continue until the whole system has a uniform temperature depending on the consistency of the heat source.

7 0

3 years ago