Calorimetry 2.
If a sample of brass (SH = 0.092cal/g°C) initially at 85.0 °C were dropped into a 50.15 g water bath initially
at 23.5 °C causing the temperature of the water to rise to 27.2 °C, what was the mass of the brass
sample? (Identify each variable below to help you set up the problem and show all equations and
work with units.)
Water

mwater =
Tiwater =
Tfwater=
AT water =
SHwater=
qwater=

Metal
mmetal =
Timetal =
Tfmetal =
ATmetal =
SHmetal =
qmetal =

Respuesta :

The mass of the brass sample is 34.89 g

We'll begin by calculating the heat gained by the water. This can be obtained as follow:

Mass of water (M) = 50.15 g

Initial temperature of water (T₁) = 23.5 °C

Final temperature of water (T₂) = 27.2 °C

Change in temperature of water (ΔT) = T₂ – T₁

Change in temperature of water (ΔT) = 27.2 – 23.5

Change in temperature of water (ΔT) = 3.7 °C

Specific heat capacity of water (C) = 1 cal/gºC  

Heat gained by water (Q) =?

Q = MCΔT

Q = 50.15 × 1 × 3.7

Q = 185.555 calories

Thus, the heat gained by the water is 185.555 calories

Next, we shall determine the heat lost by the metal. This can be obtained as follow:

Heat lost by metal = Heat gained by water

Heat gained by water = 185.555 calories

Thus,

Heat lost by metal = Heat gained by water = 185.555 calories

Recall:

Heat lost is always negative.

Therefore,

Heat lost by metal = –185.555 calories

Finally, we shall determine the mass of the brass sample. This can be obtained as follow:

Specific heat capacity of metal (C) = 0.092 cal/g°C

Initial temperature of metal (T₁) = 85.0 °C

Final temperature of metal (T₂) = 27.2 °C

Change in temperature of metal (ΔT) = T₂ – T₁

Change in temperature of meta (ΔT) = 27.2 – 85

Change in temperature of metal (ΔT) = –57.8 °C  

Heat lost by metal = –185.555 calories

Mass of brass (M) =?

Q = MCΔT

–185.555 = M × 0.092 × –57.8

–185.555 = –5.3176M

Divide both side by –5.3176

[tex]M =\frac{-185.555}{-5.3176}[/tex]

Mass of metal = 34.89 g

Therefore, the mass of the brass sample is 34.89 g

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