Question

100 g of a mystery metal at 250 C is added to 400 mL of water at 15 C. The final temperature of the system is 15.9 C. What is the heat capacity

of the mystery metal?

(PLEASE HELP need the answer by 10/21/20
11:00pm)

Answer 1

c = 0.064 j/g.°C

Step-by-step explanation:

Given data:

Mass of metal = 100 g

Initial temperature of metal = 250°C

Volume of water = 400 mL

Initial temperature of water = 15°C

Final temperature = 15.9°C

Heat capacity of metal = ?

Solution:

First of all we will calculate the heat lost by metal into water.

ΔT= 15.9°C - 15°C

ΔT= 0.9 °C

q = mcΔT

q = 400 g× 4.184 j/g.°C × 0.9 °C

q = 1506.24 j

Specific heat capacity of meta;

ΔT = 15.9°C - 250°C

ΔT = -234.1 °C

q = mcΔT

-1506.24 j = 100 g× c × -234.1 °C

(negative sign with q value indicate heat is lost by metal)

-1506.24 j = -23410 g.°C × c

c = -1506.24 j /-23410 g.°C

c = 0.064 j/g.°C