Some very hot rocks have a temperature of $260 ^o C$ and a specific heat of $20 J/(Kg*K)$ . The rocks are bathed in $40 L$ of boiling water. If the heat of the rocks completely vaporizes the water, what is the minimum combined mass of the rocks?

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Answer 1 · 2018-03-16T06:45:01

The mass of the rocks is $=28212.5kg$

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

For the rocks $DeltaT_o=260-100=160^@$

Heat of vaporisation of water is

$H_w=2257kJkg^-1$

The specific heat of the rocks is

$C_o=0.02kJkg^-1K^-1$

Volume of water is $V=40L$

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

The mass of water is $m_w=Vrho=40kg$ ,

$m_o C_o (DeltaT_o) = m_w H_w$

$m_o*0.02*160=40*2257$

The mass of the rocks is

$m_o=(40*2257)/(0.02*160)$

$=28212.5kg$

Some very hot rocks have a temperature of 260 ^o C260 ^o C and a specific heat of 20 J/(Kg*K)20 J/(Kg*K). The rocks are bathed in 40 L40 L of boiling water. If the heat of the rocks completely vaporizes the water, what is the minimum combined mass of the rocks?