Question

The amount of heat energy required to raise the temperature of water of mass 500g from 20degrees celicius to 25 degrees Celsius​

Answer 1

$\;\;\underline{\textbf{\textsf{ Given:-}}}$⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• Mass of water, m = 500 g = 0.5 kg

( For converting :- 1 kg = 1000 gm)

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•Initial temperature, T₁ = 20° C

•Final temperature, T₂= 25° C

$\;\;\underline{\textbf{\textsf{ To Find :-}}}$

• What's the mount of heat energy required to raise the temperature of water?

$\;\;\underline{\textbf{\textsf{ Solution :-}}}$

$\underline{\:\textsf{We know that :-}}$

$\underline{\boxed{ \sf s = \dfrac{Q}{m (T_2 - T_1)} }}$

Where,

• Specific heat capacity of water,

s = 4200 J/kg C

$\underline{\:\textsf{Now, substitute the given values in the formula :-}}$

$\\\\ \longrightarrow \sf 4200 = \dfrac{Q}{0.5 (25 - 20)} \\\\ \longrightarrow \sf 4200 = \dfrac{Q}{0.5 \times 5} \\\\ \longrightarrow \sf 4200 = \dfrac{Q}{2.5} \\\\ \longrightarrow \sf Q = 4200 \times 2.5 \\\\ \longrightarrow \sf Q = 10,500 J \\\\ \ \longrightarrow \underline{\boxed{\sf{ Q = 10.5 KJ }}}$

$\;\;\underline{\textbf{\textsf{ Hence-}}}$

⠀

$\underline{\textsf{ Amount of heat energy required \textbf{10.5 KJ }}}.$

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Answer 2

⟶ Given :-

• Mass of water, m = 500 g = 0.5 kg
• InitiaI temperature, T₁ = 20° C
• FinaI temperature, T₂ = 25° C

⟶ To find :-

The Amount of heat energy required

⟶ Solution :-

we know,

Specific heat capacity of water (s) = 4200 J/kg C

then,

$\Longrightarrow\sf s = \dfrac{Q}{m(T_2 - T_1)}$

$\Longrightarrow\sf 4200= \dfrac{Q}{0.5(25 - 20)}$

$\Longrightarrow\sf 4200 = \dfrac{Q}{0.5 \times 5}$

$\Longrightarrow\sf 4200 = \dfrac{Q}{2.5}$

$\Longrightarrow\sf Q = 4200 \times 2.5$

$\Longrightarrow\sf Q = 10500J$

$\Longrightarrow\boxed {\sf Q = 10.5KJ}$

Hence, the amount of heat energy required is 10.5KJ