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snow_tiger [21]
2 years ago
7

A 2 kg object has a specific heat capacity of 1,700 J/(kg \cdot⋅oC)

Physics
1 answer:
Nutka1998 [239]2 years ago
8 0

The amount of heat needed to raise the temperature of a 2kg object from 15°C to 25°C is 34000J.

HOW TO CALCULATE SPECIFIC HEAT CAPACITY:

  • The amount of heat absorbed by an object can be calculated by using the following expression:

  • Q = m.c.∆T

Where;

  1. Q = amount of heat absorbed or released (J)
  2. m = mass of object
  3. c = specific heat capacity (J/g°C)
  4. ∆T = change in temperature (°C)

  • According to this question, 2 kg object has a specific heat capacity of 1,700J/kg°C and was raised from a temperature of 15 Celsius to 25 Celsius. The heat absorbed is calculated as follows:

  • Q = 2 × 1700 × {25 - 15}

  • Q = 3400 × 10

  • Q = 34000J

  • Therefore, the amount of heat needed to raise the temperature of a 2kg object from 15°C to 25°C is 34000J.

Learn more about how to calculate heat absorbed at: brainly.com/question/11194034?referrer=searchResults

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{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Focal\:length=10\:cm}

\:\:\:\:\bullet\:\:\:\sf{Object \ distance = -15\:cm}

\\

{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Nature \: of \:the\:image}

\\

{\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\

<h3>☯ <u>By using formula of Lens</u> </h3>

\\

\dashrightarrow\:\: {\boxed{\sf{\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}}}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v}-\dfrac{1}{-15}=\dfrac{1}{10}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{10}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{30}}

\\

\dashrightarrow\:\: \sf{ v = 30 \ cm}

\\

<h3>☯ <u>Now, Finding the magnification </u></h3>

\\

\dashrightarrow\:\: \sf{ m = \dfrac{-30}{-15}}

\\

\dashrightarrow\:\: \sf{m = -2}

\\

<h3>☯ <u>Hence</u>,\\</h3>

\:\:\:\:\star\:\:\:\sf{Image \ distance = 30 \ cm}

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