A125-g metal block at a temperature of 93.2 °c was immersed in 100. g of water at 18.3 °c. given the specific heat of the metal ( s = 0.900 j/g·°c), find the final temperature of the block and the water

34.17°C

Explanation:

Given:

mass of metal block = 125 g

initial temperature = 93.2°C

We know

  ..................(1)

Q= Quantity of heat

m = mass of the substance

c = specific heat capacity

c = 4.19 for H₂O in

= change in temperature

Now

The heat lost by metal = The heat gained by the metal

Heat lost by metal =

Heat gained by the water =

thus, we have

=

⇒

Therefore, the final temperature will be = 34.17°C

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