Calculating the Required Heat: Heating Water in an Aluminum Pan

The pan and the water are always at the same temperature. When you put the pan on the stove, the temperature of the water and the pan is increased by the same amount. We use the equation for heat transfer for the given temperature change and masses of water and aluminum. The specific heat values for water and aluminum are given in the previous table.

Because the water is in thermal contact with the aluminum, the pan and the water are at the same temperature.

1. Calculate the temperature difference.
   $$\Delta T=T_f-T_i=60.0\text{°C}$$

   11.8
2. Calculate the mass of water using the relationship between density, mass, and volume. Density is mass per unit volume, or $\rho =\frac{m}{V}$ . Rearranging this equation, solve for the mass of water.
   $$m_w=\rho \cdot V=1000{\text{ kg/m}}^{\text{3}}\text{×}\left(\text{0}\text{.250 L×}\frac{\text{0}{\text{.001 m}}^{\text{3}}}{\text{1 L}}\right)\text{=0}\text{.250 kg}$$

   11.9
3. Calculate the heat transferred to the water. Use the specific heat of water in the previous table.
   $$Q_w=m_w{c}_w\Delta T=\left(0.250\text{ kg}\right)\left(\text{4186 J/kg}\text{°C}\right)\left(\text{60}\text{.0}\text{°C}\right)\text{ = 62}\text{.8 kJ}$$

   11.10
4. Calculate the heat transferred to the aluminum. Use the specific heat for aluminum in the previous table.
   $$Q_{Al}=m_{Al}{c}_{Al}\Delta T=\left(0.500\text{ kg}\right)\left(\text{900 J/kg}\text{°C}\right)\left(\text{60}\text{.0}\text{°C}\right)\text{ = 27}{\text{.0 ×10}}^3\text{J = 27}\text{.0 kJ}$$

   11.11
5. Find the total transferred heat.
   $$Q_{Total}=Q_w+Q_{Al}\text{= 62}\text{.8 kJ + 27}\text{.0 kJ = 89}\text{.8 kJ}$$

   11.12

The percentage of heat going into heating the pan is

The percentage of heat going into heating the water is