Question

Part B

The amount of boiling water required to raise the temperature of 25.0 kgof water in the bath to body temperature is 4.80 kg.

In this process, the heat lost by the boiling water is equal to the heat gained by the room-temperature water. How much heat was transferred in this process?

Express your answer to four significant figures and include the appropriate units.

n this problem answers are requested to three significant digits for grading purposes. The true number of significant digits may be more or less.

Part A

A volume of 95.0 mL of H2O is initially at room temperature (22.00 ?C). A chilled steel rod at 2.00 ?C is placed in the water. If the final temperature of the system is 21.50  ?C , what is the mass of the steel bar?

Use the following values:

specific heat of water = 4.18 J/(g??C)

specific heat of steel = 0.452 J/(g??C)

Express your answer to three significant figures and include the appropriate units. mass of the steel =

Answer 1

Part B)

Here,

Mass of water in the room temperature = 4.8 Kg

Mass of boiling water = 25 Kg

Initial temperature of boiling water = 100 ^oC

Initial temperature of room temperature water = 25 ^oC

By assuming that the final temperature of the system is T ^oC

Heat lost by the boiling water is equal to the heat gained by the room temperature water

Therefore,

4.8 Kg x(100-T) ^oC= 25 Kg x (T-25) ^oC

T = [100 + (25 x 5.2)]/6.2 = 37 ^oC

Therefore,

Heat transferred in the process

q = (25 x 10³ g) x (4.184 J/g^oC) x (37-25) ^oC = 1255 KJ

Part A)

Here,

Mass of water = 95 g

Initial temperature = 22 ^oC

Final temperature = 21.5 ^oC

Here, water is losing heat and steel is gaining heat

Therefore,

Temperature change(T) = (21.5 -22) ^oC = -0.5 ^oC

-qwater = +qsteel

By substituting the values

-95g x (4.18 J/g^oC) x (21.5-22) ^oC = m x (0.452 J/g^oC) x (21.5-2) ^oC

-95g x (4.18 J/g^oC) x (-0.5) ^oC = m x (0.452 J/g^oC) x 19.5 ^oC

198.55 = m x 8.814

m = 198.55/ 8.814 = 22.5 g