Some very hot rocks have a temperature of #560 ^o C# and a specific heat of #120 J/(Kg*K)#. The rocks are bathed in #24 L# of boiling water. If the heat of the rocks completely vaporizes the water, what is the minimum combined mass of the rocks?

Question

Some very hot rocks have a temperature of #560 ^o C# and a specific heat of #120  J/(Kg*K)#.  The rocks are bathed in #24 L# of boiling water.  If the heat of the rocks completely vaporizes the water, what is the minimum combined mass of the rocks?

Elena Albarran · Accepted Answer

The mass of the rocks is #=981.3kg#

The heat transferred from the hot rocks to the water, is equal to the heat used to evaporate the water.

For the rocks #DeltaT_o=560-100=460^@#

Heat of vaporisation of water is

#H_w=2257kJkg^-1#

The specific heat of the rocks is

#C_o=0.120kJkg^-1K^-1#

Volume of water is #V=24L#

Density of water is #rho=1kgL^-1#

The mass of water is #m_w=Vrho=24kg#,

# m_o C_o (DeltaT_o) = m_w H_w #

#m_o*0.120*460=24*2257#

The mass of the rocks is

#m_o=(24*2257)/(0.120*460)#

#=981.3kg#

Nicholas Agrusa · Answer

undefined To find the minimum combined mass of the rocks, you can use the formula:

$$Q_{\text{rocks}} = m_{\text{rocks}} \cdot c_{\text{rocks}} \cdot \Delta T_{\text{rocks}}$$

$$Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}} + m_{\text{water}} \cdot L_v$$

Given:
$Q_{\text{rocks}} = Q_{\text{water}}$,
$c_{\text{rocks}} = 120 \, \text{J/(kg*K)}$,
$\Delta T_{\text{rocks}} = 560^\circ C$,
$c_{\text{water}} = 4186 \, \text{J/(kg*K)}$,
$\Delta T_{\text{water}} = 100^\circ C$,
$L_v = 2260 \, \text{kJ/kg}$,
$V_{\text{water}} = 24 \, \text{L}$.

First, calculate the heat energy required to vaporize the water, then use that to find the mass of the rocks.