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A student measures that 81,500 J of thermal energy were added to 0.5 kg of water. If the specific heat of water is 4,184 J/kg 0C, what was its change of temperature?

Respuesta :

Answer:

[tex]\Delta T=38.95^{\circ} C[/tex]

Explanation:

Given that,

Heat measured, Q = 81500 J

Mass of water, m = 0.5 kg

The specific heat of water is 4,184 J/kg °C

We need to find the change in temperature. The heat measured is given by :

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

Where

[tex]\Delta T[/tex] is the change in temperature

[tex]\Delat T=\dfrac{Q}{mc}\\\\\Delat T=\dfrac{81500}{0.5\times 4184 }\\\\\Delta T=38.95^{\circ} C[/tex]

So, the change in temperature is [tex]38.95^{\circ} C[/tex].

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