Question

A 42.77 g sample of a substance is initially at 23.2 °C. After absorbing 248.8 cal of heat, the temperature of the substance is 188.3 °C. What is the specific heat (c) of the substance?

Answer 1

how should we relate the heat absorbed to the specific heat and the temperature change?

Answer 2

I know that heat(cal)/mass(g)xtemperature change=cal/(gxC)
I did put everything in place but got the wrong answer I don't know what I did wrong
248.8g/42.77x(188.3-23.2C)=cal/(gxC)
which = to 0.0452 cal/(gxC) did I do something wrong?

Answer 3

248.8 cal / ( 42.77 grams * (188.3 - 23.2 deg C) ) = .03523 cal per gram per deg C

Answer 4

how did you get this result , I did every stage right but still got it wrong

Answer 5

assuming i got it right, i think the calculator is not interpreting your instructions correctly.

Answer 6

I tried it again , I got the same answer as you, how would you round it off since I think there should be 3 significant figures. is it 0.300?

Answer 7

i actually think there should be 4 sig fig. because the temperature difference 165.1 has four sig figs, as does the other measurements.

Answer 8

Ok , so that would be 0.3000 as a final answer right?

Answer 9

i would only mess with the fourth significant number, not the second, third, and fourth numbers, so i would say it's .03523 or 3.523 10^-2

Answer 10

ok thank you