Question

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current attempt in progress
water at 21.0 °c is sprayed onto 0.167 kg of molten gold at 1063 °c (its melting point). the water boils away, forming steam at 100.0 °c and leaving solid gold at 1063 °c. what is the minimum mass of water that must be used?
$m_w = $

Explanation:

Step1: Calculate heat lost by gold

The heat lost by gold \(Q_{gold}=m_{gold}L_f\), where \(m_{gold} = 0.167\space kg\) and \(L_f=6.30\times 10^{4}\space J/kg\) (latent heat of fusion of gold).
\(Q_{gold}=0.167\times6.30\times 10^{4}\)

Step2: Calculate heat gained by water

The heat gained by water has two parts: heating water from \(T_1 = 21.0^{\circ}C\) to \(T_2=100.0^{\circ}C\) and vaporizing water.
The heat for heating water \(Q_{heat}=m_wc_w\Delta T\), where \(c_w = 4186\space J/(kg\cdot^{\circ}C)\) and \(\Delta T=100 - 21=79^{\circ}C\)
The heat for vaporizing water \(Q_{vap}=m_wL_v\), where \(L_v = 2.26\times 10^{6}\space J/kg\)
So \(Q_{water}=m_w(4186\times79 + 2.26\times 10^{6})\)

Step3: Equate heat lost and heat gained

Since \(Q_{gold}=Q_{water}\)
\(0.167\times6.30\times 10^{4}=m_w(4186\times79 + 2.26\times 10^{6})\)
First, calculate \(4186\times79=330694\)
\(4186\times79 + 2.26\times 10^{6}=330694+2260000 = 2590694\)
Then \(m_w=\frac{0.167\times6.30\times 10^{4}}{2590694}\)

Answer:

\(m_w\approx0.00405\space kg\)