Respuesta :

Given:
m = 325 g, the mass of water
ΔT = 77 - 15 = 62 °C, the change in temperature

Note that
c = 4.184 J/(g-°C), the specific heat of water.

Use the formula
Q = m*c*ΔT
where Q is the heat required for the change in temperature.

Q = (325 g)(4.184 J/g-°C)*(62 °C) = 8.43 x 10⁴ J
That is,
Q = 84.3 x 10³ J = 84.3 kJ

Answer:  84.3 kJ

The Quantity of heat required to raise the temperature of 325 grams of water form 15.0 °C to 77 °C is  84.308 kilojoules.

Further Explanation:

Heat capacity  

  • Heat capacity refers to the amount of heat that is required to raise the temperature of an object or a substance by one degree Celsius or 1 Kelvin.
  • Heat capacity is thus expressed as Joules/°C

Specific heat capacity  

  • This is the amount of heat that is required to raise the temperature of a unit mass of a substance by 1 degree Celsius or 1 Kelvin.
  • It is expressed as Joule/kg/°C or J/g/°C. For example the specific heat of water is 4.184 j/g/°C

Quantity of Heat  

  • Quantity of Heat is calculated using the formula;

Q = mCΔT

  • Where Q is the Quantity of heat, m is the mass of the object, C is the specific heat capacity of the object and ΔT is the resulting change in temperature.
  • ΔT is given by the difference between initial temperature and final temperature.  

That is; ΔT = Tfinal - Tinitial

In our case;

Mass of water = 325 g

Specific heat capacity of water = 4.184 J/g/°C

ΔT = 77 °C - 15.0°C

     = 62 °C

But: Q = mcΔT

Therefore;

                    Q = 325 g x 4.184 J/g/°C x 62 °C

                       =84307.6 Joules

                       = 84.308 kiloJoules

Keywords: Change in temperature, Specific heat capacity, Quantity of heat

Learn more about;

  • Specific heat capacity: https://brainly.com/question/4759369
  • Heat capacity: https://brainly.com/question/4759369
  • Quantity of heat calculation example: https://brainly.com/question/4759369

Level: High school

Subject; Physics

Topic: Quantity of Heat

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