Water at 233C is sprayed onto 0172 kg of molten gold at 1063

Water at 23.3°C is sprayed onto 0.172 kg of molten gold at 1063°C (its melting point). The water boils away, forming steam at 100°C and leaving solid gold at 1063°C. What is the minimum mass of water that must be used?

m_w = _________

Solution

Heat lost by gold,

Q1 = mL

Latent heat of fusion of gold, L = 63000 J / kg

Q1 = 0.172 * 63000 = 10836 J

Let, mass of water = M

Heat gain by water,

Q2 = MCdT + ML

Q2 = M*4186*(100 - 23) + M*2260.4*10^3

Q2 = 2.58*10^6*M J

From the principal of calorimetry,

Q1 = Q2

10836 = 2.58*10^6*M

M = 0.00419 kg

M = 4.195 g