An object with a mass of #4 kg#, temperature of #240 ^oC#, and a specific heat of #15 J/(kg*K)# is dropped into a container with #15L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

Question

An object with a mass of #4 kg#, temperature of #240  ^oC#, and a specific heat of #15 J/(kg*K)# is dropped into a container with #15L # of water at #0^oC #.  Does the water evaporate?  If not, by how much does the water's temperature change?

Olivia Askew · Accepted Answer

The object does not vaporise the water. It causes a temperature change of #0.23^@"C"# in the water.

Evaporation of a liquid can occur at any temperature. So we could say "Yes" in answer to the question "Does the water evaporate?" However, let’s assume the question meant "Does all of the water get vaporised?" (I.e. is the water raised to its boiling point and then boiled away?) To answer the question I will structure my solution as follows: Some preliminary data and calculations. Conversion of the volume of water into m³: #V = 15 L = 15 × 10^(-3) m^3# Calculation of the mass of water: #m_w = ρV = 1000 × 15 × 10^(-3) = 15 kg# Specific heat capacity of water: #c_w = 4200 J.kg^(-1).K^(-1)# Latent heat of vaporisation of water: #L_(v(water)) = 334 kJ.kg^(-1)# An initial note about the question. The provided data indicates that the object has a smaller mass and a much lower specific heat capacity than water. Thus, it appears that the object will have a very small impact on the water, so small in fact that I wonder if some of the data that was provided was inaccurate. 1. The amount of energy needed to raise water's temperature to boiling. #DeltaQ_1 = m_wc_wDeltaθ = 15 × 4200 × (100 - 0)# #⇒ DeltaQ_1 = 6.3 MJ# 2. The amount of energy needed to evaporate all the water. #DeltaQ_2 = m_wL_(v(water)) = 15 × 334 × 10^3# #⇒ DeltaQ_2 = 5.0 MJ# 3. Energy that the 4 kg object can provide. The lowest temperature the object could fall to and successfully vaporise the water would be #100^@"C"#. So the maximum temperature change would be from #240^@"C"# to #100^@"C"#. Data for the object: #m_o = 4 kg#, #c_o = 15 J.kg^(-1).K^(-1)# #DeltaQ_3 = m_oc_oDeltaθ = 4 × 15 × (240 - 100) = 8400 J# Compare the energies. We can see that 8400 J is much less than 6.4 MJ (#∆Q_1#). So the object is unable to heat the water to anywhere near its boiling point. It is not even necessary to compare 8400 J with the energy needed to evaporate the water (which was determined in part 2). 4. How much does the temperature of the water change? We just need to apply the specific heat capacity equation for the object and the water because we already know that the latter does not go through a phase change. I will define the final temperature of the water as #θ_x#. Since all energy transfers stop when both the object and the water have the same temperature the object’s final temperature will also be #θ_x#. Object starts at #240^@"C"# and ends at #θ_x#. Water starts at #0^@"C"# and ends at #θ_x#. We can say the following if we assume that the water absorbs all of the heat energy that the object transfers: #DeltaQ_w = DeltaQ_o# #⇒ m_wc_wDeltaθ_w = m_oc_oDeltaθ_o# ① Taking note of the temperature variations mentioned above, we can further define the terms ∆θ: #Deltaθ_w = θ_x – 0# #Deltaθ_o = 240 – θ_x# Replace those with those in equation ①: #⇒ m_wc_w(θ_x – 0) = m_oc_o(240 – θ_x)# #⇒ m_wc_wθ_x = 240m_oc_o – m_oc_oθ_x# Rearrange for #θ_x#: #⇒ θ_x (m_wc_w + m_oc_o) = 240m_oc_o# #⇒ θ_x = (240m_oc_o)/(m_wc_w + m_oc_o)# Now enter the values into the formula: #⇒ θ_x = (240 × 4 × 15)/(15 × 4200 + 4 × 15) = 0.23^@"C"# So the temperature change in the water is #0.23^@"C"#, i.e. #0^@"C"# to #0.23^@"C"#.

William Audy · Answer

undefined To determine if the water evaporates, we need additional information, such as the latent heat of vaporization. Assuming no evaporation, the water's temperature increases by approximately 10.67 degrees Celsius.