tokanotfound
contestada

4000 J of heat energy is applied to a 70 g sample of water initially at 35oC. What is the final temperature of the sample?
Please with the solution

Respuesta :

samuelonum1

Answer:

48.7°C

Explanation:

Step one:

given data

Quantity of heat = 4000J

mass of sample= 70g= 0.07kg

initial temperature T1=35°C

Water has a specific heat capacity of 4182 J/kg°C

Required

The final temperature T2

Step two:

Q= mcΔT

4000= 0.07*4182(T2-35)

4000=292.74(T2-35)

4000=292.74T2-10245.9

collect like terms

4000+10245.9=292.74T2

14245.9= 292.74T2

divide both sides by 292.74

T2= 14245.9/ 292.74

T2=48.7°C

Cricetus

The final temperature of the sample will be "48.7°C". To understand the calculation, check below.

Heat and temperaure

According to the question,

Heat quantity, Q = 4000 J

Sample mass, m = 70 g or,

                       = 0.07 kg

Initial temperature, T₁ = 35°C

Specific heat capacity, c = 4182 J/kg°C

We know the relation,

→    Q = mcΔT

By substituting the values, we get

4000 = 0.07 × 4182 (T₂ - 35)

         = 292.74 (T₂ - 35)

         = 292 T₂ - 10245.9

Now,

4000 + 10245.9 = 292.74 T₂

            14245.9 = 292.74 T₂

                      T₂ = [tex]\frac{14245.9}{292.74}[/tex]

                           = 48.7°C

Thus the above answer is correct.    

Find out more information about temperature here:

https://brainly.com/question/224374

Otras preguntas