A 82.5g sample of a certain metal at 99.0c is added to 150.0g sample of water at 21.0c. After thermal equilibrium is established, the final temperature of the mixture is 27.5c. Assuming no heat is lost to the surroundings or the calorimeter, what is the specific heat of the metal?

Question

jfraser · Answer

What's the equation that deals with heat, mass, and change in temp?

anonymous · Answer

would it be q=m(delta)T?

jfraser · Answer

it is Q = mCDT. Since the heat (Q) is being lost by the metal, it must be gained by the water, so the 2 objects must have equal Q's$$Q_{metal} = m_{metal}*C_{metal}* \Delta  T_{metal} = Q_{water} = m_{water}*C_{water}* \Delta T_{water}$$ this means we don't actually need to solve for Q, just set the 2 of them equal to each other. This makes the only unknown piece the C of the metal. Rearrange and solve for C of the metal.$$C_{metal} = \frac{m_{water}*C_{water}* \Delta T_{water}}{m_{metal}*\Delta T_{metal}}$$