# Bomb Calorimetry: Solve for Water Temp.

• Mitchtwitchita
In summary: I'm having a real tough time with this question:If the specific heat of the ball bearing is 0.474 J/g x C, calculate the final temperature of the water. Assume the calorimeter to have negligible heat capacity.The ball bearing heats up the water until the temperature of the ball is equal to the temperature of the water. The final temperature of the water is 21.27 degrees Celsius.
Mitchtwitchita
Hey guys, I'm having a real tough time with this question:

A 30.14-g stainless steel ball bearing at 117.82 degrees C is place in a constant pressure calorimeter containing 120.0 mL of water at 18.44 degrees C. If the specific heat of the ball bearing is 0.474 J/g x C, calculate the final temperature of the water. Assume the calorimeter to have negligible heat capacity.

qsteel +qwater = 0
qsteel = -qwater

qsteel = ms(deltaT)
=(30.14 g)(0.474 J/g x C)(18.44 degrees C - 117.82 degrees C)
=-1420 J

Therefore qwater =1420 J
1420 J = (120 g)(4.184 J/g x C)(Tfinal - 18.44 degree C)
Tfinal = 1420/502 + 18.44
=21.27

However, the answer in the book is 21.19 degrees celsius. I don't think that I'm doing this one properly. Can anybody set me straight?

I think that going off the assumption that $$\Delta T = T_{ball} - T_{water}$$ might be where the deviation is arising, remember, the ball is heating up a limited amount of water up until a point when $$T_{ball} = T_{water}$$, your case would be true if there was a massive amount of water relative to the ball, because the final temperature of the system would be pretty close to the waters original temperature anyway.

I would go about finding the final temperature as:
$$q_{steel} = mC(x-117.82)$$
$$q_{water} = mC(x-18.44)$$

Since the change in heat energy must be equal to zero as you stated you find:
$$q_{steel}= -q_{water}$$
which is what you stated

Therefore:
$$mC_{steel}(x-117.82) = -mC_{water}(x-18.44)$$

Rearranging:
$$mC_{steel}x - mC_{steel}\times 117.82 = mC_{water}\times 18.44 - mC_{water}x$$

$$mC_{steel}x+mC_{water}x = mC_{water}\times 18.44 + mC_{steel}\times 117.82$$

$$x(mC_{steel}+mC_{water}) = mC_{water}\times 18.44 + mC_{steel}\times 117.82$$

$$x = \frac{mC_{water}\times 18.44 + mC_{steel}\times 117.82}{mC_{steel}+mC_{water} }$$

Substituting:

$$x = \frac{120\times 4.184\times 18.44 + 30.14\times 0.474\times 117.82}{30.14\times 0.474+120\times 4.184} = 21.1896^{\circ}C$$

Thanks again AbedeuS!

## 1. What is bomb calorimetry?

Bomb calorimetry is a laboratory technique used to measure the energy released from a chemical reaction. It involves placing a sample of the substance in a sealed container (bomb) surrounded by water, and then measuring the change in temperature of the water as the substance is burned.

## 2. How is bomb calorimetry used to solve for water temperature?

In bomb calorimetry, the energy released from a chemical reaction is directly proportional to the change in temperature of the surrounding water. By measuring the change in temperature and using mathematical equations, the initial temperature of the water can be solved for.

## 3. What type of substances can be analyzed using bomb calorimetry?

Bomb calorimetry is commonly used to analyze organic compounds such as food, fuels, and other biological materials. However, it can also be used to analyze inorganic compounds under certain conditions.

## 4. How accurate is bomb calorimetry in determining the water temperature?

The accuracy of bomb calorimetry depends on various factors such as the precision of the equipment, the heat capacity of the water and the sample, and the skill of the operator. With proper calibration and technique, bomb calorimetry can provide highly accurate results.

## 5. What are the advantages of using bomb calorimetry over other methods of measuring energy?

Bomb calorimetry is a direct and precise method of measuring the energy released from a chemical reaction, as it eliminates the need for external factors such as air or pressure. It also allows for the analysis of a wide range of substances and can provide both qualitative and quantitative data.

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