Question

50.0 grams of copper, with a specific heat capacity of 0.385 J/goC, and the initial temperature of 10.0 oC is dropped into 400.0 grams of vegetable oil, with a specific heat capacity of 1.67 J/goC at 175 oC. Calculate the equilibrium temperature of the oil and copper combination, assuming no heat is absorbed from or by the container, or the surroundings, (Show all work clearly, including formulas, units, etc.).

Answer 1

Answer:

final temperature T2 = 35.17 °C

Explanation:

given in attachment

Answer 2

Given:

Mass of Cu= 50g

Specific heat capacity of Cu= 0.385 J/g°C

Initial temp of Cu= T₁ = 10°C

Mass of oil= 400g

Specific heat capacity of oil= 1.67J/g°C

Initial temp of oil= T₂= 175°C

To find:

The equilibrium temperature of the mixture (T')

Solution:

Since given that no heat is exchanged with the container or the surroundings

⇒ Heat lost by Oil= Heat gained by Cu

till a common temperature T' is achieved

We know that Heat (Q) = mCΔT

where m = mass of the object

C = specific heat capacity

ΔT= Final Temp - Initial Temp

According to the question,   $Q_{Oil} = - Q_{Cu}$

or 400 X 1.67 X (T' - 175) = - [50 X 0.385 (T' - 10)]

solving,

668 x T' - 668 X 175 = -(19.25 T' - 192.5)

or 668T' - 116900 = - 19.25T' + 192.5

or 687.25T' = 117092.5

or T' = 170.37 °C

Hence, the equilibrium temperature of the mixture is 170.37 °C.