Specifc Heat of Material?

Question

The specific heat of a material can be used to identify it. For example, a 100.0 g sample of a substance is heated to 100.0°C and placed into a calorimeter cup (having a negligible amount of heat absorption) containing 150.0 g of water at 25°C. The sample raises the temperature of the water to 32.1°C. With the information, identify the substance.

I used two formulas.
Cmetal = Qgained by the water / (mass of metal)(change in temp of metal) WHERE Qgained by the water = (mass of water)(change in temp of metal)(4180 J/kg · K).

I first converted all masses to terms of "kg" and "Kelvin" first.

Qgained by the water = (0.15 kg)(305.25 K - 298.15 K)(4180 J/kg · K)
= 4451.7 J
Cmetal = 4451.7 J / (0.10 kg)(305.25 K - 373.15 K)
= 655 J/kg · K ???

The problem is I got a big specific heat, 655 J/kg · K, but it is probably wrong. Where did I make the mistake? I couldn't identify the substance, what is it really supposed to be? Thank you for your help.

Answer 1

I see nothing wrong with your method. The specific heat you got isn't particularly high; compare your answer (655 J/kg-K) to the specific heat of water (4180 J/kg-K). If your answer isn't listed in a particular table you're looking at, make sure your units match; specific heat is often listed as J/g-K or J/mol-K, or sometimes even with calories instead of Joules (shudder). To me, it looks like your answer is close to calcium.

Answer 2

Your process looks correct to me; I don't think you made a mistake. The specific heat you got is not unusually high. I couldn't find a material to match it, either, but this specfic heat falls between those of iron (450 J/kg*K) and aluminum (897 J/kg*K). You might find it on a more complete list of metals and their specific heat capacities, if your textbook has one.