How much heat is required to melt 45.3g of liquid cesium from 85.9 C to 120.2 C? (specific heat of liquid cesium=0.252J/gC)

Question

asked Dec 6, 2023 156k views

1 Answer

TTZ · Answer 1 · 2023-12-10T06:06:45+0000

391.56 Joules

The formula for calculating the amount of heat required is expressed as:

$Q=mc\triangle t$

where:

• m is the mass = 45.3g

,

• specific heat of caesium c = 0.252J/gC

,

• change in temperature △t = 120.2 - 85.9 = 34.3 degrees celsius

Substitute the given parameters into the formula

$\begin{gathered} Q=45.3g*(0.252J)/(g^oC)*34.3^0C \\ Q=391.56Joules \end{gathered}$

Hence the amount of heat required is 391.56 Joules