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!

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!

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Answer 2

Amelia Adams

To find the minimum combined mass of the rocks, we first need to calculate the heat energy required to vaporize the water using the formula:

Q = mcΔT

where:
Q = heat energy
m = mass
c = specific heat capacity
ΔT = change in temperature

Given:
Initial temperature of water = 100°C
Final temperature of water = 100°C (as it completely vaporizes)
Specific heat of water = 4186 J/(Kg*K)

Using the formula, we can calculate the heat energy required to vaporize the water:

Q_water = mcΔT
Q_water = (45 kg)(4186 J/(Kg*K))(100°C - 100°C)
Q_water = 0 J

Now, the heat energy from the hot rocks will vaporize the water. The heat energy released by the hot rocks is given by:

Q_rocks = mcΔT
Q_rocks = (m_rocks)(210 J/(Kg*K))(420°C - 100°C)
Q_rocks = 120300m_rocks J

Since the heat energy from the rocks must completely vaporize the water, we have:

Q_water = Q_rocks
0 J = 120300m_rocks J

Therefore, the minimum combined mass of the rocks is 0 kg, as the heat energy from the rocks is not sufficient to vaporize the water.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.