Question

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?

Answer 1

Answer:

$\Delta T=38.95^{\circ} C$

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 :

$Q=mc\Delta T$

Where

$\Delta T$ is the change in temperature

$\Delat T=\dfrac{Q}{mc}\\\\\Delat T=\dfrac{81500}{0.5\times 4184 }\\\\\Delta T=38.95^{\circ} C$

So, the change in temperature is $38.95^{\circ} C$.