A 25-g sample of an alloy at 91 degrees celsius is placed in 50g water at 20 degrees celsius. If the final temperature reached by the alloy sample and water is 27 degrees celsius, what is the specific heat (cal/g degree celsius) of the alloy?

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jfraser · Answer

take a look at this question, from earlier. http://openstudy.com/users/jfraser#/updates/51129ad1e4b07c1a5a646728 The equation for heat transfer is $$Q = m*C_P * \Delta T$$ In this case there's not one thing transferring energy, there's 2: the metal alloy, and the water. The second law of thermo says that whatever energy the alloy loses must be gained by the water, in exactly equal amounts, so: $$m_{water} *C_{water} *\Delta T_{water} = Q_{water} = Q_{alloy} = m_{alloy}*C_{alloy}*\Delta T_{alloy}$$removing the Q's from the middle gets $$m_{water} *C_{water} *\Delta T_{water} = m_{alloy}*C_{alloy}*\Delta T_{alloy}$$ of those 6 pieces, you already know five. Rearrange and solve for C_alloy