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

Answer:

$0.52 \frac{J}{g~^{\circ}C}$

Explanation:

The to start with the equation:

$Q=m~Cp$ΔT

Where:

Q= Heat (J)

m= mass (in grams)

Cp= Specific heat ($\frac{J}{g~^{\circ}C}$)

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":

$Cp=\frac{89.7~J}{30.0~g*5.20^{\circ}C}=0.52 \frac{J}{g~^{\circ}C}$

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