Question

What is the specific heat of titanium in j/(g⋅∘c) if it takes 89.7 j to raise the temperature of a 33.0 g block by 5.20 ∘c?

Answer 1

`0.52  (J)/(g~^(\circ)C)`

Step-by-step explanation:

The to start with the equation:

`Q=m~Cp`ΔT

Where:

Q= Heat (J)

m= mass (in grams)

Cp= Specific heat (
`(J)/(g~^(\circ)C)`)

ΔT=Tfinal-Tinitial

Then we have to put the values into the equation:

`89.7~J=33.0~g*Cp*5.20^(\circ)C`

The next step would be the to solve for "Cp":

`Cp=(89.7~J)/(30.0~g*5.20^(\circ)C)=0.52  (J)/(g~^(\circ)C)`

Answer 2

Specific heat capacity is the amount of energy required to raise one gram of substances by 1 degree celsius . Therefore specific heat capacity for tatanium is 89.7j /( 33.0g x5.2 degree celsius) = 0.52j/g degree celcius
Molar mass for tatanium is 47.9 g/mole
heat is therefore 47.9 g/mole x 0.52j/g =24.9j/mole