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 : The change in temperature will be, [tex]22.89^oC[/tex]
Explanation :
First we have to determine the mass of water.
[tex]Density=\frac{Mass}{Volume}[/tex]
Given :
Density of water = 1 g/mL
Volume of water = 355 mL
[tex]1.00g/mL=\frac{Mass}{355mL}[/tex]
[tex]Mass=355g[/tex]
Now we have to determine the change in temperature.
Formula used :
[tex]Q=m\times c\times \Delta T[/tex]
where,
Q = heat absorb = 34 kJ = 34000 J (1 kJ = 1000 J)
m = mass of water = 355 g
c = specific heat of water = [tex]4.184J/g^oC[/tex]
[tex]\Delta T[/tex] = change in temperature = ?
Now put all the given value in the above formula, we get:
[tex]34000J=355g\times 4.184J/g^oC\times \Delta T[/tex]
[tex]\Delta T=22.89^oC[/tex]
Therefore, the change in temperature will be, [tex]22.89^oC[/tex]
