Question

A 400-lb piece of aluminum at 700°F is dropped into 500 lb of water at 70°F. What is the final temperature? Show working equation. Answer must be in English units

Answer 1

here,

the mass of alumunium piece , m1 = 400 lb = 181.4 kg

the initial temperature of alumunium piece , T1 = 700 F = 371.11 degree C

the mass of water , m2 = 500 lb = 226.8 kg

the initial temperature of water , T2 = 70 F = 21.11 degree C

let the final temperature be Tf

using Principle of Calorimetry

m1 * Ca * ( T1 - Tf) = m2 * Cw * ( Tf - T2)

181.4 * 900 * ( 371.11 - Tf) = 226.8 * 4186 * ( Tf - 21.11)

solving for Tf

Tf = 72.5 degree C = 162.5 F

the equivalent temperature is 162.5 F