## Specific heat /heat capacity :Examples of calculations

(We consider isolated systems, uneven otherwise stated, so the no warmth is lost to the environment and also the principle of preservation of energy can be used to work out the answer.

You are watching: How many kilocalories are needed to raise 2kg of water 50 degrees celsius

)

revise the theory>>

1) How much heat is essential to advanced the temperature of a block of copper ( through mass 0.5 kg) indigenous 0°C come 100° C ? (for copper, c = 386 J / kg oC)

2) How much heat is required to raise the temperature of 0.5 kg of water native 0°C to 100° C? (for water, c = 4186 J / kg oC)

3)What would certainly be the final temperature of a mixture that 100 g of water at 90°C and also 600 g of water at 20°C ?

4)What would be the final temperature if a 2 kg item of lead at 200°C is put in a container with 10 kg that water in ~ 50°C ? (for lead, c = 128 J / kg oC)

5) 2 equal recipients A and B, v 2 various liquids, at first at 20°C, room heated through a warm plated and both receive the exact same amount of heat. As a result, the temperature of fluid A is raised to 40°C and also that of liquid B is elevated to 80°C. If the liquids are gotten rid of from their recipients and also mixed, the final temperature would be around: a) 45°C b) 50°C c) 55°C d) 60°C e) 65°C

6) A heat exchanger consists of a coil shame pipe wraped roughly a broader pipe.The water in the coil is offered to cool the water on the main pipe and also it has a circulation rate that 18 together /s ; the enters at 20ºC and also leaves at 40 ºC (as shown on the image). The water ~ above the main pipe has actually a flow rate of 18 l/s. If that enters at 85ºC, what will be its departure temperature?

a) 75 ºC b) 65 ºC c) 55 ºC d) 45 ºC e) 35 ºC Examples including melting ice are a bit an ext tricky and can be discovered here: practice involving also latent heat>>

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

1) using the formula: Q = 386 * 0.5 * 100 = 19300 J or 19.3 kJ

Comments: that is necessary to observe the SI units. The fixed is in kg and also the heat energy in J. Usually the temperature is converted into K, but because we are taking the distinction (or the variation), the doesn"t matter what systems are used (if kelvin or celsius). However, if the difference of temperatures have the right to be plugged in the formula in the scales kelvin or celsius (no need to transform celsius into kelvin). If the difference is in degrees farenheit a switch (to celsius or kelvin) is needed prior to using the formula.

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

Comments: note that this is much more than 1o times the nergy required in the case of copper!

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

We understand that heat flows from the hotter body to the cooler body. Hence, the water in ~ the higher temperature will certainly "loose" heat and the water at the lower initial temperature will "gain" heat. The correct means of explicate this instance is by saying that warmth will be transferred from the hotter to the cooler water.

We also know that, by preservation of energy, the quantity of heat lost will be the same that is gained.

So, let"s speak to the last temperature of the mixture Tf.

The lot of warmth that will be moved from the hotter water is:

4186 * 0.1 * (90 - Tf)

The quantity of heat that will certainly be transferred to the cooler water is:

4186 * 0.6 * ( Tf -20)

Because these two quantities need to be equal, we have an equation:

4186 * 0.1 * (90 - Tf) = 4186*0.6 * (Tf - 20)

We need to find Tf:

418.6 * (90-Tf) = 2511.6 * (Tf-20)

Getting rid that the brackets:

37674 - 418.6 Tf = 2511.6 Tf - 50232

-2930.2 Tf = -87906

Tf = 30° C

4) comparable to 3) above,

in this case:

amount of warmth transferred native the lead:

128*2*(200-Tf)

amount of heat transferred to the water:

4186*10*(Tf-50)

Equating the 2 heats:

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

256(200-Tf)=41860(Tf-50)

51200-256 Tf = 41860 Tf - 2093000

42116 Tf = 2144200

Tf = 50.9°

5) This is a very an excellent problem and also I will carry out an prolonged answer.

Initial considerations:

Both liquids obtain the very same amount the heat yet liquid B achieves a greater temperature. That method that fluid A requires much more heat in order come raise the tempearature. In various other words, fluid A has a certain heat the is greater than the particular heat o liquid B.

Then, the heat source is removed and also the liquids are mixed. What is the last temperature? that is temptaing come guess 60 ºC due to the fact that that is half way between the 2; that would certainly be correct if both liquids had actually the same specific heat, but that is no the situation as we know the A has actually a greater specific heat. Because of that, the heat transferred from fluid B (that is hotter) can not raise the temperature the A very much.

So, this intial considerations allow us to remove some answers. The final temperature should be lower than 60 ºC.

Calculations

-Parte 1

Both liquids obtain the very same amount of warmth so QA= QB:

QA = cA mA ΔT = cA mA (40-20) = cA mA*20

QB = cB mB ΔT = cB mB (80-20) = cB mB*60

Equating both formulas:

-Parte 2

The last temperature (Tf) completed when the liquids are combined up need to be intermediate between the two. The name is one will certainly lose warm that will certainly be got by the other. This quantity of heat must be equal because no warm is shed to the environment and energy need to conserve. Based on that we can write one more equation:

(80 - Tf ) cB mB = (Tf-40) cA mA

Using (I):

(80 - Tf ) cB mB = (Tf-40) 3 cB mB

Tf = 50 ºC

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

6) This warmth exchanger calculation is frequently performed in chemistry or mechanical design practice.

See more: In A Specific Ecosystem, The Population With The Greatest Number Of Members Will Be The

Let´s consider what happens throughout 1 min:

-Part 1

We speak to Q1 the heat received by the water in the coil. The certain heat of water is stood for by c and the density of water is 1 kg/l together usual.

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

Q1 = 18 *c *20 = 360 c

-Part 2

We speak to Q2 the warmth lost by the mater ~ above the key pipe. The is cooled from 85ºC

to a last temperature Tf i m sorry we must calculate:

Q2 = 12 * c * (85 - Tf)

-Part 3

The heat lost by the water in the tube need to equal the heat gained by the water in the coil, therefore taht:

Q1 = Q2

360 c = 12 (85 - Tf) c

Cancelling c:

360 = 1020 - 12 Tf

-660 = -12 Tf

Tf = 55 ºC

OBS:The particular heat the water to be not provided because it would certainly be cancelled anyway.