help!! really hard question?

Question

the specific heat of ice is 0.48cal/g degree Celsius. how much heat will it take to raise temp. of 10 g of ice frm -30 C to -20 C?

a. 0.48 cal

b. 48 cal

c. 10 cal

d. 4.8 cal

Answer 1

-30C to -20C is a10degC difference.

10gIce x 10degC x 0.48cal/g-degC = 48 cal (b.)

Answer 2

I think it's best to get an understanding by doing this problem in full-sentence, paragraph form. Explaining it in spoken words with a chalkboard would be even better, but what the heck.

The units on the specific heat tell you, straight out, that for every ONE gram of water, you'll need 0.48 calories to increase the temperature by ONE degree celsius.

First, let's take one gram up ten degrees: we need 0.48 calories for each degree for this one gram. This means to raise our single gram by one degree we need 0.48 calories of heat input. So to raise our single gram ten degrees, it will take ten times that much, or 4.8 calories.

Now lets raise ten grams ten degrees. We know we need 4.8 calories of energy input for each gram to go up ten degrees, so we can intuit that we need ten times that much energy input for ten grams to go up ten degrees, or 48 calories.

Your answer is B, like the man said.

Does that make sense? Did I say it well?

Write the problem down on a piece of paper and do the problem forwards and backwards to prove to yourself that the logic works. Practice by making ten grams go up one degree first, then use that value to make ten grams go up ten degrees. You'll get the same two answers.

Answer 3

d

i think

we learned about this but im not sure

u like subtract the change (-30 - - 20) and get - 10 and then divide by .48?

wait no

idk

i m like 100% sure that c is wrong thou
and a is wrong cuz it didnt change

i m gonna guess
B

Answer 4

its 10 cal

Answer 5

C

Answer 6

C.