Apiece of metal weighing 59.047 g was heated to 100.0 °c and then put it into 100.0 ml of water (initially at 23.7 °c). the metal and water were allowed to come to an equilibrium temperature, determined to be 27.8 °c. assuming no heat lost to the environment, calculate the specific heat of the metal.

Law of conservation of energy

Heat Loss by the metal  = Heat Gain by the water  (Assume no heat loss to the environment)

M₁C₁ΔT₁= M₂C₂ΔT₂

Where

C₁ and C₂ = the specific heat capacities of the metal and water = C₁?, 4.182J/gK

ΔT₁ and ΔT₂ = Temperature changes of the metal and the water

ΔT₁ = (373 - 300.8)K and ΔT₂= (300.8 - 296.7)K   convert to S.I units

M₁ and M₂ are the masses of the metal and water = 59.047g, but

M₂ = density x volume = 1kg/m³ x 0.1m³=0.1kg

Note: Volume of water = 100ml=0.1m³, also density of water = 1kg/m³

We have,

0.059kg x C₁J/KgK (373 - 300.8)K = 0.1kg x 4182J/kgK (300.8 - 296.7)K

4.259C₁=1715J/KgK

C₁=402.5J/KgK

The specific heat of the metal is 402.5J/KgK

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form