Question

Some very hot rocks have a temperature of `420 ^o C` and a specific heat of `210 J/(Kg*K)`. The rocks are bathed in `45 L` of boiling water. If the heat of the rocks completely vaporizes the water, what is the minimum combined mass of the rocks?

Answer 1

`M_r = (1.02xx10^8)/(320*210)~~1513kg`
You will need lots of rocks or one big boulder...roughly with dimensions of `[80xx80xx80 ->90xx90xx90] cm` Good Luck!

Explanation:

Given or Known:
`V_w=45L; c_w= 4180J/(kg*""^0C);`
`H_v=2.26xx10^6J/(kg), T_f=100^oC`

`T_r=420^oC; C_r=210 J/(kg*""^0C)` K or C is the same
Unknown: Heat required vaporize ice and
`Q_v; Q_(liquid)= Q_w; Q_R, and M_r`

Required: `color(brown)(M_r)`?

Calculate: Heat Required to vaporized the water
`M_w= 45kg; Q_v = M_w*H_v`
`Q_v= 45kg* 2.26xx10^6 J/(kg) ~~ 1.02xx10^8 J`

Calculate: the heat gained by liquid water
At boiling point all the, further addition of energy causes the substance to undergo vaporization. All the added thermal energy converts the substance from the liquid state to the gaseous state. The temperature does not rise while a liquid boils.
So `Q_(liquid)=0` by virtue that `DeltaT = 0` stays for water stays at `100^o` during the vaporization process.

Calculate: the heat loss by the rock as it cools down to `100^oC`
`Q_r= M_rc_r(T_i-T_f)= M_r*210 J/(kg*""^0C)(420-100)^oC=`
`Q_r=210*320M_r`

Now the heat loss by rock above is used up to vaporize the 45 kg of water. At thermal equilibrium `Q_R= Q_v+Q_w`; with `Q_w=0`
`Q_r=Q_v`
`210*3208M_r= 1.02xx10^8; M_r = (1.02xx10^8)/(320*210)~~1513kg`
You will need lots of rocks or one big one...roughly with dimensions of `.8xx.8xx.8 ->.9xx.9xx.9` Good Luck!