Question

Suppose that you warm up $\mathbf{500}\mathbf{g}$of water (half a litre, or about a pint) on a stove while doing$\mathbf{\text{5×10J}}$of work on the water with an electric beater. The temperature of the water is observed to rise from${\mathbf{\text{20}}}^{\mathbf{\text{o}}}\mathbf{\text{C}}$to${\mathbf{\text{80}}}^{\mathbf{\text{o}}}\mathbf{\text{C}}$. What was the change in the thermal energy of the water? Taking the water as the system, how much transfer of energy$\mathbf{Q}$due to a temperature difference was there across the system boundary? What was the energy change of the surroundings?

Short answer

For the amount of water heated the following observations are made:

Change in thermal energy of the water -${\mathrm{\Delta E}}_{\mathrm{thermal}}=1.26\times {10}^5\mathrm{J}$

Transfer of energy due to a temperature difference is$\mathrm{Q}=7.6\times {10}^4\mathrm{J}$

Energy change in the surroundings is${\mathrm{\Delta E}}_{\text{surroundings}}=-1.26\times {10}^5\text{J}$

Step-by-step solution

Define the energy principle

The free energy principle defines how non-equilibrium steady-states are maintained in living and non-living systems by restricting their states to a minimal number. It establishes that systems minimise a free energy function of their internal states, implying hidden states in the environment.

Calculation of change in energy of water

Use the definition of thermal energy in order to obtain the change of energy in the water $\text{Δ}{\mathrm{E}}_{\text{thermal}}$.

$\begin{array}{c}{\mathrm{\Delta E}}_{\text{thermal}}={\left(\mathrm{mC\Delta T}\right)}_{\text{H2O}}\\ =\left(\text{500g}\right)\left(\text{4.2}\frac{\text{J}}{{\text{g}}^{\text{o}}\text{C}}\right)\left({\text{80}}^{\text{o}}{\text{C-20}}^{\text{o}}\text{C}\right)\\ ={\text{1.26×10}}^{\text{5}}\text{J}\end{array}$

Therefore, the change in energy of water is obtained as $\text{Δ}{\mathrm{E}}_{\text{thermal}}{\text{=1.26×10}}^5\text{J}$.

Equation for heat

According to the energy principle, this change of energy in the water is equal to the input of heat and work. Write this into an equation, in which then proceed to solve for the heat $\mathrm{Q}$.

${\mathrm{\Delta E}}_{\mathrm{thermal}}=\mathrm{Q}+\mathrm{W}$

Solve for$\mathrm{Q}$ and substitute the values,

$\begin{array}{c}\mathrm{Q}={\mathrm{\Delta E}}_{\mathrm{thermal}}-\mathrm{W}\\ {\text{=1.26×10}}^{\text{5}}{\text{J-5×10}}^{\text{4}}\text{J}\\ {\text{=7.6×10}}^{\text{4}}\text{J}\end{array}$

Therefore, the equation of heat is obtained as$\text{Q}={\text{7.6×10}}^{\text{4}}\text{J}$ .

Energy change in surroundings 

The energy gained by the mass of water is equal to the energy loss of the surroundings by considering the surroundings as everything but the mass of water.

Hence, the electric beater is also part of the surroundings.

Write the relation between energy of surrounding and thermal energy.

$\begin{array}{c}{\mathrm{\Delta E}}_{\text{surroundings}}=-{\mathrm{\Delta E}}_{\text{thermal}}\\ =-{\text{1.26×10}}^{\text{5}}\text{J}\end{array}$

Therefore, energy change in the surroundings is obtained as$-{\text{1.26×10}}^{\text{5}}\text{J}$ .