You find it takes 200 J of heat to take 4 kg of an unknown substance from 200 K to 240 K. It does not change phases during this interval. If you added 300 J of heat to the substance instead of 200 J, what would be its final temperature?

300 K

Explanation:

First, we have find the specific heat capacity of the unknown substance.

The heat gained by the substance is given by the formula:

H = m*c*(T2 - T1)

Where m = mass of the substance

c = specific heat capacity

T2 = final temperature

T1 = initial temperature

From the question:

H = 200J

m = 4 kg

T1 = 200K

T2 = 240 K

Therefore:

200 = 4 * c * (240 - 200)

200 = 4 * c * 40

200 = 160 * c

c = 200/160

c = 1.25 J/kgK

The heat capacity of the substance is 1.25 J/kgK.

If 300 J of heat is added, the new heat becomes 500 J.

Hence, we need to find the final temperature, T2, when heat is 500 J.

Using the same formula:

500 = 4 * 1.25 * (T2 - 200)

500 = 5 * (T2 - 200)

100 = T2 - 200

=> T2 = 100 + 200 = 300 K

The new final temperature of the unknown substance is 300K.

v 154.9= 0.9 length length s little over 154.9cm

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2