Question

A 74.0-gram piece of metal at 94.0 °C is placed in 120.0 g of water in a calorimeter at 26.5 °C. The final temperature in the calorimeter is 32.0 °C. Determine the specific heat of the metal. Show your work by listing various steps, and explain how the law of conservation of energy applies to this situation.
** I know someone already answered this on another post, but I think their answer is incorrect** I did it and think I have the right answer, but I want to double check it. Thanks!

Answer 1

Answer:

Explanation:

Heat lost by metal = mass x specific heat x fall in temperature

= 74 x S x ( 94 - 32 )

= 4588 S

heat gained by water = mass x specific heat x rise in temperature

= 120 x 1 x ( 32 - 26.5 )   ( specific heat of water is 1 cals / gm )

= 660

Heat lost = heat gained

4588S = 660

S = .14 cal /gm .

Specific heat of metal = .14 cal  / gm