(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Suppose molten (liquid) lead, mass ml = 9.83 kg, is at its melting point. The lead is poured into water of mass mw = 585 g and initial temperature Tw = 16.2 C. Find mwb, the amount of the water that boils.

NOTE: Reaching the boiling point is not enough; the question asks for the amount of water that vaporizes as well.

ASSUME: no water evaporates; it only vaporizes at the boiling point.

2. Relevant equations

Constants: M.P Lead = 327 C, MP Water = 100 C

Latent Heat Lead = 33 kJ/kg

Latent Heat Q = mL

Specific Heat Eq'n: Q=mcΔT

3. The attempt at a solution

mL + mcΔT = mcΔT + mL

9.83 kg (33 kJ/kg) + 9.83 kg(0.130 kJ/kg)(327C - 100C) = 0.585(4.186)(100 -16.2) + m(2260 kJ/kg)

Solving for m, we get m = 0.215 kg.

Did I go wrong in my reasoning here? Many thanks in advance for your help!

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# Homework Help: Latent Heat + Specific Heat Problem

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