A copper pot with a mass of 0.500 kg contains 0.170 kg of water, and both are at 20.0°C. A 0.250-kg block of iron at 85.0°C is dropped into the pot. Find the final temperature of the system, assuming no heat loss to the surroundings.

Verified step by step guidance

1. Identify the specific heat capacities for copper, water, and iron. The specific heat capacity is the amount of heat required to change the temperature of one kilogram of a substance by one degree Celsius. For copper, it is approximately 0.385 J/g°C, for water it is 4.186 J/g°C, and for iron, it is about 0.449 J/g°C.
2. Convert the masses of the copper pot, water, and iron block from kilograms to grams because the specific heat capacities are given in J/g°C.
3. Set up the heat transfer equation assuming no heat loss to the surroundings. The heat gained by the copper pot and water should equal the heat lost by the iron block. Use the formula: $m_c{c}_c(T_f-T_{c_i})+m_w{c}_w(T_f-T_{w_i})=m_i{c}_i(T_{i_i}-T_f)$, where $m$ is mass, $c$ is specific heat capacity, $T_f$ is the final temperature, and subscripts $c$, $w$, and $i$ refer to copper, water, and iron respectively.
4. Solve the equation for the final temperature $T_f$. This involves combining like terms and isolating $T_f$ on one side of the equation.
5. Check the physical plausibility of your answer, ensuring that the final temperature is between the initial temperatures of the substances involved (between 20.0°C and 85.0°C) and that the direction of heat flow is from the hotter iron to the cooler water and copper pot.