A piece of metal is heated by placing it in hot oil. It is removed from the hot oil and dropped into a beaker of cold water. The water heats up due to the transfer of heat from the metal. What happens to the temperature of the beaker? How is the heat that causes the temperature of the glass to rise accounted for in calorimetry

a) The temperature of the beaker rises as this transfer of heat goes on.

b) Check Explanation.

Explanation:

a) The heat lost by the piece of metal is normally gained by the all the components that it comes in contact with after the heating procedure.

(Heat lost by piece of metal) = (Heat gained by the cold water) + (Heat gained by the beaker).

So, since heat is also gained by the Beaker, its temperature should rise under normal conditions.

That is essentially what the zeroth law of thermodynamics about thermal equilibrium talks about.

If two bodies are at thermal equilibrium with reach other and body 2 is in thermal equilibrium with a third body, then body 1 and body 3 are also in thermal equilibrium

Temperature of the piece of metal decreases, temperature of water rises and the temperature of the beaker rises as they all try to attain thermal equilibrium.

b) In calorimetry, the aim is usually for the water (in this case) to take up all of the heat supplied by the piece of metal. Hence, the calorimeter is usually heavily insulated (or properly called lagged). Thereby, reducing the amount of heat that the calorimeter would gain.

But in cases where the heat lost to the insulated calorimeter isn't negligible, the heat capacity of the calorimeter is usually obtained and included it is included in the heat transfer calculations.

Hope this Helps!!!