A metal with a specific heat of 0.70 cal/g℃ and a mass of 8.00g absorbs 48.0cal of heat. What will the temperature change of the metal be?

Question

Having some trouble with this particular problem so if anyone is able to assist me, that would be great!

Answer 1

This problem requires a good understanding of significant figures and a proper use of the following formula:

Q = mcΔT ; where Q = the quantity of heat absorbed = 48.0 cal ; m = mass = 8.00 g ;

ΔT = temperature change ; and c = specific heat = 0.70 cal/g℃

Plugging the values above into the formula gives us :

48.0 cal = (8.00 g)( 0.70 cal/g℃)ΔT

48.0 cal = (5.6 cal/℃)ΔT

To solve for ΔT, divide BOTH sides of this equation by 5.6 cal/℃;

SO: (48.0 cal) / (5.6 cal/℃) = (5.6 cal/℃)ΔT / (5.6 cal/℃)

This gives us: 8.57 ℃ = ΔT

However, since the specific heat value has only 2 significant figures, and we are multiplying and dividing, the answer must have only 2 significant figures.

So, we must round the Final Answer to 2 significant figures, giving the Final Answer as

ΔT = 8.6 ℃