Frank M. answered • 11/07/14

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This is a problem in specific heats of different materials. Let

- m
_{w}= the mass of the water & m_{s}= the mass of the metal. - c
_{w }= the specific heat of water, which is 4.18J/g * K & c_{s}= the specific heat of the sample, which is unknown. - The final temperature of the water = T
_{f.} - The initial temperature of the water = T
_{i. }Then - Since the temp of the water rises, the delta T for it is T
_{f }- T_{i, }or 5 degrees Kelvin. - Since the temp of the sample falls, its delta T is T
_{i}- T_{f}, or (95 - 51.6) degrees Kelvin. Then - m
_{w}c_{w}(delta T_{w}) = m_{s}c_{s}(delta T_{s}). Solve for c_{s}to get the specific heat of the metal sample. Evaluate the expression left of the equal sign to obtain the quantity of heat transferred from the metal to the water. - Additional hint: one ml of water weighs one gram.