An object with a mass of #4 kg#, temperature of #350 ^oC#, and a specific heat of #6 J/(kg*K)# is dropped into a container with #28 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

Answer 1

No. The water increases from #0^oC to 13.9^oC.#

To calculate an actual final water temperature we equate the two thermodynamic changes and solve for the common value of temperature. Water specific heat is 4.178 J/g-oK. Water mass can be taken as 1.0g/cm^3 for this calculation. Heat available from the object: 4kg * 6J/kg⋅K * delta #T^oK#

To heat the water, 28000g * 4.178 J/g-oK = 116984 J/oK is needed, which is equal to 28L * 1000cm^3/L * 1.0g/cm^3 = 28000g of water.

116984 J/oK * (T-0)#T^oK# = 24J/⋅K * (350-T)#T^oK#

116984T J = 8400 - 24T ; 117008*T = 8400 ; deltaT = 13.9 So, the water increases from #0^oC to 13.9^oC#, it does not evaporate.

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Answer 2

Julia Aguilar

No, the water does not evaporate. The temperature of the water will increase by approximately 30.8°C.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.