A74.0-gram piece of metal at 94.0 °c is placed in 120.0 g of water in a calorimeter at 26.5 °c. the final temperature in the calorimeter is 32.0 °c. determine the specific heat of the metal. show your work by listing various steps, and explain how the law of conservation of energy applies to this situation.

Explanation:

The guiding principle in calorimetry is the Law of Conservation of Energy: the sum of all the energy transfers must add up to zero. That is,

q₁ + q₂ + … = 0

The formula for the heat q gained or lost by a substance is

q = mCΔT

where

m = the mass of the substance.

 C = its specific heat capacity.

ΔT = the change in temperature.

In this problem, there are two heat transfers: the heat lost by the metal and the heat gained by the water.

m₁C₁ΔT₁ + m₂C₂ΔT₂ = 0

Data:

m₁ = 74.0 g;   T₁ = 94.0 °C; T₂ = 32.0 °C

m₂ = 120.0 g; T₁ = 26.5 °C; T₂ = 32.0 °C

Calculations:

ΔT₁ = 32.0 - 94.0 = -62.0 °C

ΔT₂ = 32.0 - 26.5 =    5.5 °C

14,16 18 reacts with group

i think its 100.

hope this !