Question

A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The calorimeter and the water have a combined mass of 250.0 g and an overall specific heat of 1.035 cal/g•°C. The initial temperature of the calorimeter is 10.00°C. The system reaches a final temperature of 11.08°C when the metal is added. Any heat lost by the metal is

Answer 1

Answer:

gained by the calorimeter. on edge

Explanation:

I just did it

Answer 2

Answer:- Heat lost by the metal is 279.45 cal.

Solution:- This type of problems are solved by using the concept, heat given = - heat taken

Metal temperature is decreasing from 45.00 degree C to 11.08 degree C. It means the heat is lost by the metal and this heat lost by metal is gained by water and the calorimeter to raise their temperature.

the equation we use is, $q=mc\Delta T$ .

where, q is the heat energy, m is mass, c is specific heat and $\Delta T$ is change in temperature.

Combined mass of calorimeter and water is 250.0 g and the specific heat is $\frac{1.035cal}{g.^0C}$ .

$\Delta T$  for calorimeter and water (combined) = 11.08 - 10.00 = 1.08 degree C

$\Delta T$  for metal = 11.08 - 45.00 = -33.92 degree C

let's plug in the values in the above equation and calculate heat gained by combined system.

q=250.0g*\frac{1.035cal}{g.^0C}*1.08^0C

q = 279.45 cal

So, the heat lost by the metal is 279.45 cal.