Question

A metal with a specific heat of 0.70 J/gºC and a mass of 8.00 g absorbs 48.0 J of heat.

What will be the temperature change of the metal? 8.57 °C

Answer 1

Answer:8.8.6°C

Step-by-step explanation:

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 ℃