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].
