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Umar has two copper pans, each containing 500cm3 of water. Pan A has a mass of 750g and pan B has a mass of 1.5kg. Which pan will require the least energy to heat the water to 100°C?

Respuesta :

Answer:

heat required in pan B is more than pan A

Explanation:

Heat required to raise the temperature of the substance is given by the formula

[tex]Q = ms\Delta T[/tex]

now we know that both pan contains same volume of water while the mass of pan is different

So here heat required to raise the temperature of water in Pan A is given as

[tex]Q_1 = (m_w s_w + m_ps_p)\delta T[/tex]

[tex]Q_1 = (0.5(4186) + 0.750(s))\Delta T[/tex]

Now similarly for other pan we have

[tex]Q_2 = (m_w s_w + m_ps_p)\delta T[/tex]

[tex]Q_2 = (0.5(4186) + 1.50(s))\Delta T[/tex]

So here by comparing the two equations we can say that heat required in pan B is more than pan A

onyebuchinnaji

The pan that will require the least energy to heat the water to 100°C is pan A.

The given parameters;

  • volume of water, V = 500 cm³
  • mass of pan A = 750 g
  • mass of pan B = 1.5 kg = 1,500 g
  • final temperature of water, t = 100 ⁰C.

The total energy required to raise the temperature of the water from the initial temperature to final temperature is calculated as follows;

[tex]Q_t = Q_{pan} \ + Q_{water}[/tex]

[tex]Q = mc\Delta t[/tex]

where;

m is the mass of each pan

The pan with more mass will require more energy while the pan with least mass will require less energy.

Thus, the pan that will require the least energy to heat the water to 100°C is pan A.

Learn more here:https://brainly.com/question/19520516

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