A 18 g sample of metal is heated to 96°C and placed in a calorimeter containing 21 g of water at an initial temperature of 24°C. After the metal cools in the water bath, the final temperature inside the calorimeter is 26°C. What is the (positive) specific heat of the metal rounded to the HUNDREDTHS place? (hint: final temperatures of metal and water are the same).

THE SPECIFIC HEAT OF THE METAL IS 0.14 J/g K

Explanation:

First, write out the variables in the question

Mass of sample = 18 g

Initial temp of metal = 96°C = 96 + 273 = 369 K

Mass of water = 21 g

Initill temp of water = 24°C = 24 + 273 K = 297 K

final temp of water and metal  = 26°C = 26 + 273 K = 299 K

temperature change = (

In calorimetry, there is conservation of energy and heat liberated by one is absorbed by the other. That is heat gained is equals to heat lost.

Heat lost by the metal = Heat gained by water

mc ΔT = mc ΔT

18 * C * ( 299 - 369 ) = 21 * 4.2 * ( 299 - 297)

18 * C * -70 = 21 * 4.2 * 2

-1260 C = 176.4

C = 176.4 / 1260

C = 0.14 J/ g K

The specific heat of the metal is hence 0.14 J/g K

it is called acid bonding

the potential energy of the reactants is 200j.

from the energy diagram, the energy of the product formed is 350j; this means that, this reaction is an endothermic reaction, because it absorbs energy from its environment.