Acalorimeter contains 124 g of water at 26.6°c. a block of metal with a mass of 26 g is heated to 95.8°c and then placed in the water in the calorimeter. after sufficient time, the temperature of the water is measured and found to be 29.0°c. calculate the heat capacity per gram of metal. assume no heat is lost to the calorimeter or the surroundings.

The heat capacity of the metal is 0.717 J/g°C

Explanation:

Step 1: Data given

Mass of water = 124.0 grams

Mass of metal = 26.0 grams

Temperature of water = 26.6 °C

Temperature of metal = 95.8 °C

Final temperature of the water = 29.0 °C

Step 2: Calculate heat capacity of metal

Qlost = -Q gained

Qmetal = -Qwater

Q = m*c*ΔT

⇒ with m = mass in grams

⇒ c= the heat capacity in J/g°C

⇒ ΔT = The change in temperature = T2 - T1 (in °C)

Qmetal = -Qwater

m(metal)*c(metal)*ΔT(metal) = -m(water)*c(water)*ΔT(water)

26 *c(metal) *(29-95.8) = -124 * 4.184 *(29-26.6)

-1736.8 * c(metal) = -1245.16

c(metal) = 0.717 J/g°C

The heat capacity of the metal is 0.717 J/g°C

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

they are both synthetic elements

explanation:

d