The Temperature of a 20-kg Block Increases by 5

The Temperature of a 20-kg Block Increases by 5.

Specific heat /estrus capacity :Examples of calculations

(We consider isolated systems, unless otherwise stated, so that no heat is lost to the environment and the principle of conservation of energy can be applied to work out the answer.)

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1)
How much rut is needed to raise the temperature of a cake of copper ( with mass 0.five kg) from 0°C to 100° C  ? (for copper, c = 386 J / kg
oC)


two)
How much heat is needed to heighten the temperature of 0.v kg of water from 0°C to 100° C? (for water, c = 4186 J / kg
oC)


iii)What would exist the final temperature of a mixture of 100 g of water at xc°C and 600 g of water at 20°C ?


four)What would exist the final temperature if a 2 kg slice of pb at 200°C is inserted in a container with ten kg of h2o at fifty°C ? (for atomic number 82, c = 128 J / kg
oC)


5)
Ii equal recipients A and B, with 2 different liquids, initially at

20°C
, are heated by a hot plated and both receive the same amount of heat. As a result, the temperature of liquid A is raised to
xl°C
and that of liquid B
is raised to 80°C.
If the liquids are removed from their recipients and mixed, the concluding temperature would be around:

a) 45°C b) l°C c) 55°C d) 60°C east) 65°C


6)
A heat exchanger consists of a whorl shaped pipe wraped around a wider pipage.The h2o in the scroll is used to cool the h2o on the principal pipage and it has a flow charge per unit of eighteen fifty /s ; it enters at 20ºC and leaves at forty ºC (as shown on the image). The water on the master pipe has a menstruation rate of 18 l/due south. If it enters at 85ºC, what volition exist its exit temperature?

a) 75 ºC b) 65 ºC c) 55 ºC d) 45 ºC e) 35 ºC


Examples involving melting ice are a scrap more tricky and tin can exist found hither: exercises involving also latent rut>>

instance – 9/11 physics- how much heat needed to weaken the towers?

Answers:

1)
Applying the formula:

Q = 386 * 0.5 * 100 = 19300 J or 19.3 kJ

Comments: It is of import to find the SI units. The mass is in kg and the heat energy in J. Normally the temperature is converted into Yard, but because we are taking the difference (or the variation), information technology doesn’t matter what units are used (if kelvin or celsius). However, if the difference of temperatures can exist plugged in the formula in the scales kelvin or celsius (no demand to convert celsius into kelvin). If the difference is in degrees farenheit a conversion (to celsius or kelvin) is needed before using the formula.


two)
Q = 4186 * 0.5 * 100 = 209300 J or 209.3 kJ

Comments: Note that this is more than 1o times the nergy needed in the case of copper!


3)
This example is besides typical and it requires some algebraical skills.

We know that oestrus flows from the hotter body to the cooler body. Hence, the h2o at the higher temperature will “loose” heat and the water at the lower initial temperature volition “gain” heat. The correct fashion of describing this situation is by saying that rut will be transferred from the hotter to the cooler water.

We also know that, past conservation of free energy, the amount of heat lost volition be the aforementioned that is gained.

So, let’s phone call the final temperature of the mixture Tf.

The amount of heat that volition exist transferred from the hotter water is:

4186 * 0.1 * (90 – Tf)

The amount of heat that will exist transferred to the cooler water is:

4186 * 0.six * ( Tf -20)

Considering these two quantities must be equal, nosotros take an equation:

4186 * 0.1 * (xc – Tf) = 4186*0.half dozen * (Tf – 20)

Nosotros need to find Tf:

418.6 * (90-Tf) = 2511.6 * (Tf-twenty)

Getting rid of the brackets:

37674 – 418.half-dozen Tf = 2511.half-dozen Tf – 50232

-2930.2 Tf = -87906

Tf = thirty° C


4)
Similar to three) higher up,

in this case:

amount of heat transferred from the lead:

128*2*(200-Tf)

amount of oestrus transferred to the h2o:

4186*10*(Tf-l)

Equating the 2 heats:

128*2*(200-Tf) = 4186*10*(Tf-50)

256(200-Tf)=41860(Tf-fifty)

51200-256 Tf = 41860 Tf – 2093000

42116 Tf = 2144200

Tf = 50.ix°


v)
This is a very good problem and I will provide an extended respond.

Initial considerations:

Both liquids receive the same amount of heat simply liquid B achieves a college temperature. That ways that liquid A requires more heat in club to raise its tempearature. In other words, liquid A has a specific heat that is higher than the specific heat o liquid B.

Then, the heat source is removed and the liquids are mixed. What is the final temperature? It is temptaing to guess 60 ºC because that is half mode betwixt the 2; that would be correct if both liquids had the aforementioned specific rut, only that is not the case as we know the A has a higher specific rut. Considering of that, the heat transferred from liquid B (that is hotter) cannot enhance the temperature of A very much.

So, these intial considerations enable united states to eliminate some answers. The final temperature must be lower than 60 ºC.

Calculations

-Parte 1

Both liquids receive the same corporeality of heat so QA= QB:

QA
= cA
mA
ΔT = cA
grandA
(twoscore-xx) = cA
mA*20

QB
= cB
chiliadB
ΔT = cB
mB
(80-20) = cB
gB*threescore

Equating both formulas:

cA
chiliadA*xx = cB
mB*60 ==> cA
mA
= three cB
1000B
(I)

-Parte 2

The final temperature (Tf) achieved when the liquids are mixed up must exist intermediate between the two. The hotter one will lose heat that will exist gained past the other. This amount of oestrus must be equal because no estrus is lost to the environment and free energy must conserve. Based on that nosotros can write another equation:

(80 – Tf ) cB
thouB
= (Tf-twoscore) cA
mA

Using (I):

(fourscore – Tf ) cB
one thousandB
= (Tf-40) 3 cB
kB

Tf = l ºC

Respond is b)

OBS:
Perceba que as massas e calores específicos dos líquidos não foram dadas porque seriam canceladas durante os cálculos.

6)
This heat exchanger calculation is typically performed in chemical or mechanical engineering practice.

Let´s consider what happens during 1 min:

-Part 1

We call Qi
the estrus received by the water in the coil. The specific heat of h2o is represented by c and the density of water is 1 kg/fifty as usual.

It is heated from 20ºC to 40ºC hence ΔT=20 :

Qane
= 18 *c *20 = 360 c

-Part 2

We telephone call Q2
the heat lost by the mater on the primary pipage. It is cooled from 85ºC

to a terminal temperature Tf which we must summate:

Qtwo
= 12 * c * (85 – Tf)

-Function three

The heat lost by the water in the tube must equal the heat gained past the h2o in the coil, and so taht:

Qane
= Q2

360 c = 12
(85 – Tf)
c

Cancelling c:

360 = 1020 – 12 Tf

-660 = -12 Tf

Tf = 55
ºC

OBS:The specific estrus of water was not provided because information technology would be cancelled anyway.

revise the theory>>

exercises involving also latent heat>>

example – 9/11 physics- how much heat needed to weaken the towers?

The Temperature of a 20-kg Block Increases by 5

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