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3 kg of water ([tex]c = 4,186 \, \text{J/kg}^\circ \text{C}[/tex]) is at an initial temperature of 10°C. If 7,700 J of heat is applied, what will be its final temperature?

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Answer :

[tex]\bold{\huge{\underline{ Solution }}}[/tex]

Given :-

  • Here , The initial temperature of 3kg of water is 10° C
  • The heat supplied to 3kg water is 7,700 J
  • The specific heat of water is 4186 J/kg ° C

To Find :-

  • We have to find the final temperature of water .

Let's Begin :-

Here, we have

  • Mass = 3 kg
  • Specific heat capacity = 4186 J/kg°C
  • Initial temperature = 10° C
  • Heat applied = 7,700J

We know that,

  • Specific heat is the heat that is required to increase the temperature of 1 kg or unit mass by 1° C or unit ° C

That is,

[tex]\bold{\red{ C = }}{\bold{\red{\dfrac{ Q}{m{\delta}T}}}}[/tex]

  • Here, Change in temperature is ΔT that is, Final temperature - Initial temperature

So,

[tex]\sf{{\delta} T = }{\sf{\dfrac{ Q}{Cm}}}[/tex]

Subsitute the required values,

[tex]\sf{ T2 - 10 = }{\sf{\dfrac{ 7700}{3{\times}4186}}}[/tex]

[tex]\sf{ T2 - 10 = }{\sf{\dfrac{ 7700}{12558}}}[/tex]

[tex]\sf{ T2 - 10 = }{\sf{\cancel{\dfrac{ 7700}{12558}}}}[/tex]

[tex]\sf{ T2 - 10 = 0.61 }[/tex]

[tex]\sf{ T2 = 0.61 + 10 }[/tex]

[tex]\bold{ T2 = 10.61 {\degree} C}[/tex]

Hence, The final temperature of 3kg if 7,700 J of heat supplied is 10.61 °C

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