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To make yourself some coffee, you put one cup of water (246 gg ) in a small pot on the stove. Part A What quantity of energy must be transferred thermally to the water to raise its temperature from 20 ∘C∘C to 100 ∘C∘C?

Answer :

Answer:

The required heat energy is [tex]\bf{19680~cal}[/tex].

Explanation:

The heat energy 'Q' required to raise the temperature of water is given by

[tex]Q = M~s~\Delta \theta[/tex]

where 'M' is the mass of water, 's' is the specific heat capacity of water and '[tex]\Delta \theta[/tex]' is the change of temperature.

Given, [tex]M = 246 gm, \Delta \theta = (100^{0}~C - 20^{0}~C) = 80^{0}~C[/tex] and we know that the specific heat capacity of water is [tex]s = 1~cal~gm^{-1}~^{0}C^{-1}[/tex].

Substituting the values in the above expression, the required heat energy is

[tex]Q = 246~gm \times 1~cal~gm^{-1}~^{0}C^{-1} \times 80^{0}~C = 19680~Cal[/tex]

boffeemadrid

The heat required to raise the temperature of water is required.

The heat required is [tex]82380.48\ \text{J}[/tex]

m = Mass of coffee = 246 g

c = Specific heat of water = 4.184 J/g C

[tex]\Delta T[/tex] = Change in temperature = [tex]100-20=80^{\circ}\text{C}[/tex]

Heat is given by

[tex]Q=mc\Delta T\\\Rightarrow Q=246\times 4.186\times 80\\\Rightarrow Q=82380.48\ \text{J}[/tex]

The heat required is [tex]82380.48\ \text{J}[/tex]

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