Question

What is the temperature change in 355 mL of water upon absorption of 34 kJ of heat? (The specific heat capacity of water is 4.184 J/(g⋅∘C).)

Answer 1

Answer : The change in temperature will be,
`22.89^oC`

Explanation :

First we have to determine the mass of water.

`Density=(Mass)/(Volume)`

Given :

Density of water = 1 g/mL

Volume of water = 355 mL

`1.00g/mL=(Mass)/(355mL)`

`Mass=355g`

Now we have to determine the change in temperature.

Formula used :

`Q=m* c* \Delta T`

where,

Q = heat absorb = 34 kJ = 34000 J (1 kJ = 1000 J)

m = mass of water = 355 g

c = specific heat of water =
`4.184J/g^oC`

`\Delta T` = change in temperature = ?

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

`34000J=355g* 4.184J/g^oC* \Delta T`

`\Delta T=22.89^oC`

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

Answer 2

The temperature change is 23 °C.

q = mCΔT

ΔT = q/(mC)

m = 355 g

∴ ΔT = (34 000 J)/(355 g × 4.184 J·°C⁻¹g⁻¹) = 23 °C

Note: The answer can have only two significant figures because that is all you gave for the amount of heat absorbed.