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* c* \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* 6.55J/g^oC* \Delta T`

`\Delta T=3265.479^oC`

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