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?

Mar 29, 2016

${M}_{r} = \frac{1.02 \times {10}^{8}}{320 \cdot 210} \approx 1513 k g$
You will need lots of rocks or one big boulder...roughly with dimensions of $\left[80 \times 80 \times 80 \to 90 \times 90 \times 90\right] c m$ Good Luck!

Explanation:

Given or Known:
V_w=45L; c_w= 4180J/(kg*""^0C);
${H}_{v} = 2.26 \times {10}^{6} \frac{J}{k g} , {T}_{f} = {100}^{o} C$

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: $\textcolor{b r o w n}{{M}_{r}}$?

Calculate: Heat Required to vaporized the water
M_w= 45kg; Q_v = M_w*H_v
${Q}_{v} = 45 k g \cdot 2.26 \times {10}^{6} \frac{J}{k g} \approx 1.02 \times {10}^{8} J$

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 ${Q}_{l i q u i d} = 0$ by virtue that $\Delta T = 0$ stays for water stays at ${100}^{o}$ during the vaporization process.

Calculate: the heat loss by the rock as it cools down to ${100}^{o} C$
Q_r= M_rc_r(T_i-T_f)= M_r*210 J/(kg*""^0C)(420-100)^oC=
${Q}_{r} = 210 \cdot 320 {M}_{r}$

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 $.8 \times .8 \times .8 \to .9 \times .9 \times .9$ Good Luck!