Solving Calorimetry Ques: Molybdenum Specific Heat

[vg19]

Hi,

I have 1 question that I am having trouble with.

1) A stick of molybdenum weighing 237g and starting at a temperature of 373Kelvin is thrust into 244g of water starting at 283Kelvin. If the final Temperature observed for the whole system is 288Kelvin, what would the specific heat of the molybdenum be? Ignore the calorimeter.

For this question I think I am suppose to use the heat lost/gained = mass x C x Temp Change formula and solve for C. However, I am confused on what values to sub in. There are 2 masses, and different temperature.

Any help is much appreciated

Thanks

 

[CartoonKid]

Think like this, the heat lost by molydenum = heat gained by water. $$m_{Mo}C_{Mo}(373-288)=m_{water}C_{water}(288-283)$$. Water's follows water's, Mo's follows Mo's.

 

[wais]

for the question! Let's break this problem down step by step. First, we need to determine the heat lost by the molybdenum stick and the heat gained by the water. We can do this by using the formula Q = mCΔT, where Q is the heat, m is the mass, C is the specific heat, and ΔT is the change in temperature.

For the molybdenum stick, we have a mass of 237g and a temperature change of 373K-288K = 85K. Plugging these values into the formula, we get Q = (237g)(C)(85K).

For the water, we have a mass of 244g and a temperature change of 288K-283K = 5K. Plugging these values into the formula, we get Q = (244g)(4.18J/gK)(5K) = 5113.6J.

Now, since energy is conserved in a closed system, we can set these two equations equal to each other and solve for C. This gives us:

(237g)(C)(85K) = 5113.6J

Solving for C, we get C = 0.243 J/gK. Therefore, the specific heat of molybdenum is 0.243 J/gK.

I hope this helps! Let me know if you have any other questions.