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Chemistry 12 Online
OpenStudy (anonymous):

A 135-g sample of a metal requires 2.50kj to change its temperature from 19.5 degrees Celsius to 100.0 degrees Celsius. What is the specific heat of this metal?

OpenStudy (anonymous):

I know you should do: 19.5-100.0=80.5 degrees Celsius but im having trouble with the rest it is not adding up to the actual answer which is 119.5 degrees Celsius

OpenStudy (jfraser):

the equation you need is \[Q = m * C_{P} * \Delta T\] You have Q (2500J), you have m (135g), and you have DT (80.5C). Rearrange to solve for Cp, and you should get 0.230J/gC. I don't know why you think you should get 119.5C.

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