Question

Given 10.8 grams of substance Y, if the substance absorbs 231 kilojoules of energy and the specific heat of the substance is 6.55 J/g·°C, what is the change in temperature?

Answer 1

Q = mxcxdelta t
or
231 = 10.8x6.55x Delta t
Delta t = 3.265 degrees celcius

Answer 2

Answer : The change in temperature will be, $3265.479^oC$

Explanation :

Formula used :

$Q=m\times c\times \Delta T$

where,

Q = heat absorb = 231 kJ = 231000 J     (1 kJ = 1000 J)

m = mass of substance Y = 10.8 g

c = specific heat of substance = $6.55J/g^oC$

$\Delta T$ = change in temperature = ?

Now put all the given value in the above formula, we get:

$231000J=10.8g\times 6.55J/g^oC\times \Delta T$

$\Delta T=3265.479^oC$

Therefore, the change in temperature will be, $3265.479^oC$