What is the specific heat capacity of a 50 gram piece of 100 degrees celsius metal that will change 400 grams of 20 degrees celsius water to 22 degrees celsius?

== == This answer is taken straight off Yahoo answers and I thought it would be helpful to "spread the wealth." here it is. change in temperature of metal,

75 - 18.3 = 56.7 'C

change in temperature of water,

18.3 - 15 = 3.3 'C

energy gained by water, assuming Cp water = 4.1813 J/g/'C

using the formula, Q = mCp(theta)

where,

Q = energy in Joules

m = mass in grams

Cp = specific heat capacity in J/g/'C

theta = change in temperature in 'C

3.3 * 150 * 4.1813 = 2.06974 kJ

energy gained by water = energy dissipated by metal

using the formula, Q = mCp(theta) and solving for Cp

Cp of metal = 2.06974 k / 56.7 *150 = 0.2434 J/g/'C