Question

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)

Answer 1

391.56 Joules

Explanations:

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