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

`\Delta T=38.95^(\circ) C`

Step-by-step 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=(Q)/(mc)\\\\\Delat T=(81500)/(0.5* 4184 )\\\\\Delta T=38.95^(\circ) C`

So, the change in temperature is
`38.95^(\circ) C`.