Answer :
Answer : The temperature when the water and pan reach thermal equilibrium short time later is, [tex]59.10^oC[/tex]
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
[tex]q_1=-q_2[/tex]
[tex]m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)[/tex]
where,
[tex]c_1[/tex] = specific heat of aluminium = [tex]0.90J/g^oC[/tex]
[tex]c_2[/tex] = specific heat of water = [tex]4.184J/g^oC[/tex]
[tex]m_1[/tex] = mass of aluminum = 0.500 kg = 500 g
[tex]m_2[/tex] = mass of water = 0.250 kg = 250 g
[tex]T_f[/tex] = final temperature of mixture = ?
[tex]T_1[/tex] = initial temperature of aluminum = [tex]150^oC[/tex]
[tex]T_2[/tex] = initial temperature of water = [tex]20^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]500g\times 0.90J/g^oC\times (T_f-150)^oC=-250g\times 4.184J/g^oC\times (T_f-20)^oC[/tex]
[tex]T_f=59.10^oC[/tex]
Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, [tex]59.10^oC[/tex]
Final answer:
In Physics, the final temperature of the water and the aluminum pan when they reach thermal equilibrium can be calculated by applying the conservation of energy and the formula Q = mcΔT using the specific heats and masses of the materials and the initial temperatures.
Explanation:
The subject of this question is Physics, and it pertains to the concept of thermal equilibrium in calorimetry. When 0.250 kg of 20.0°C water is poured into a 0.500-kg aluminum pan at 150°C, heat transfer will occur between the pan and the water until they reach the same temperature. This problem involves using the principle of the conservation of energy where the heat lost by the aluminum pan equals the heat gained by the water. The specific heat capacities of water and aluminum, along with the masses and initial temperatures, will be used to calculate the final equilibrium temperature.
To solve the problem, we apply the formula Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. For water at equilibrium temperature T, the heat gained will be mwatercwater(T - 20.0°C). For the aluminum pan, the heat lost will be mpancpan(150°C - T). Setting the heat lost by the pan equal to the heat gained by the water and solving for T will give the final temperature at equilibrium.
