A 28.4 g sample of an unknown metal is heated to 39.4 °C, then is placed in a calorimeter containing 50.0 g of water. Temperature of water increases from 21.00 °C to 23.00 °C. What is the specific heat of this metal in this problem?

THE SPECIFIC HEAT OF THE METAL IS 0.8983 J/g °C

Explanation:

In solving the problem, we have to understand that:

Heat lost by the metal = Heat gained by the water in the bomb calorimeter

First is to calculate the heat evolved from the reaction

Heat = mass * specific heat * change in temperature

Mass of water = 50 g

specific heat of water = 4.184 J/g °C

Change in temperature = 23 - 21 = 2 °C

So therefore,

Heat = 50 * 4.184 * 2

Heat = 418.4 J

Next is to solve for the specific heat of the metal;

Heat lost by the metal is the same as the heat gained by water

Heat = mass * specific heat of  metal * change in temperature

Change in temperature = 39.4 °C - 23 °C = 16.4 °C

418.4 = 28.4 * C * 16.4

C = 418.4 / 28.4 * 16.4

C = 418.4 / 465.76

C = 0.8983 J/ g °C

The specific heat of the metal is hence 0.8983 J/g °C

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.