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

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.

Answer 2

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

Explanation :

First we have to determine the mass of water.

$Density=\frac{Mass}{Volume}$

Given :

Density of water = 1 g/mL

Volume of water = 355 mL

$1.00g/mL=\frac{Mass}{355mL}$

$Mass=355g$

Now we have to determine the change in temperature.

Formula used :

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

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:

$34000J=355g\times 4.184J/g^oC\times \Delta T$

$\Delta T=22.89^oC$

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