The energy given off by 300 grams of an alloy as it cools through 50°C raises the temperature of 300 grams of water from 30°C to 40°C. The specific heat of the alloy (in cal/g C ) is:

Question

i know Q=CpM(delta T)

but what is what?

Responses

* physics - bobpursley, Monday, June 22, 2009 at 8:22pm

Heat lost=heat gained
300*c*50=300*4.18*10

I did this, but i got .836, and the answer is suppose to be .2 so im not sure what i'm doing wrong!!

Answer 1

This is a simple math problem .

c=300/300 * 4.18* 10/50 = .8 j/gC about.

Now if you do specific heat in calories, you get .2. Numbers without units are meaningless.

Answer 2

Do in Cals, not Joules so 300*c*50= 300*1*10, which will give you 3000/15000= .2