Question

Suppose you apply a flame and heat one liter of water, raising its temperature 10°C. If you transfer the same heat energy to two liters, how much will the temperature rise? For three liters? Record your answers on the blanks in the drawing at the right.

Answer 1

(a) the water density is $d=1000 kg/m^3$, and 1 liter corresponds to a volume of $V=1 L=0.001 m^3$. Therefore we can find the mass of the water in the first case:
$m=dV=(1000 kg/m^3)(0.001 m^3)=1 kg$

The amount of heat supplied to the water to raise its temperature by $\Delta T=10 ^{\circ} C$ is
$Q=m C_s \Delta T$ (1)
where 
$C_s = 4.18 kJ/(kg ^{\circ} C}$ is the specific heat capacity of the water. 
Using the data, we find
$Q=(1 kg)(4.18 kJ/(kg ^{\circ} C)(10^{\circ} C)=41.8 kJ$

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
$m=dV=(1000 kg/m^3)(0.002 m^3)=2 kg$
And so by re-arranging equation (1) we can calculate the new increase of temperature:
$\Delta T_2 = \frac{Q}{m C_s} = \frac{41.8 kJ}{(2 kg)(4.18 kJ/(kg ^{\circ} C)}=5 ^{\circ} C$

(b) Now we have 3 liters of water. SImilarly to point (a), the mass is now
$m=dV=(1000 kg/m^3)(0.003 m^3)=3 kg$
And so, the increase in temperature if we use the same amount of heat as before is
$\Delta T_3= \frac{Q}{m C_s} = \frac{41.8 kJ}{(3 kg)(4.18 kJ/(kg ^{\circ} C)}=3.3 ^{\circ} C$