(a) the water density is [tex]d=1000 kg/m^3[/tex], and 1 liter corresponds to a volume of [tex]V=1 L=0.001 m^3[/tex]. Therefore we can find the mass of the water in the first case:
[tex]m=dV=(1000 kg/m^3)(0.001 m^3)=1 kg[/tex]
The amount of heat supplied to the water to raise its temperature by [tex]\Delta T=10 ^{\circ} C[/tex] is
[tex]Q=m C_s \Delta T[/tex] (1)
where
[tex]C_s = 4.18 kJ/(kg ^{\circ} C}[/tex] is the specific heat capacity of the water.
Using the data, we find
[tex]Q=(1 kg)(4.18 kJ/(kg ^{\circ} C)(10^{\circ} C)=41.8 kJ[/tex]
We want to find the increase in temperature if we transfer the same amount of heat Q to 2 liters of water. The mass of 2 liters of water is
[tex]m=dV=(1000 kg/m^3)(0.002 m^3)=2 kg[/tex]
And so by re-arranging equation (1) we can calculate the new increase of temperature:
[tex]\Delta T_2 = \frac{Q}{m C_s} = \frac{41.8 kJ}{(2 kg)(4.18 kJ/(kg ^{\circ} C)}=5 ^{\circ} C [/tex]
(b) Now we have 3 liters of water. SImilarly to point (a), the mass is now
[tex]m=dV=(1000 kg/m^3)(0.003 m^3)=3 kg[/tex]
And so, the increase in temperature if we use the same amount of heat as before is
[tex]\Delta T_3= \frac{Q}{m C_s} = \frac{41.8 kJ}{(3 kg)(4.18 kJ/(kg ^{\circ} C)}=3.3 ^{\circ} C [/tex]
