If it takes 200 j to raise the temperature of a 10-gram block of ice by 10°c, what is the specific heat of the ice?

The specific heat of ice is 2 j/g. °C

Explanation:

Given data:

Amount of heat = 200 J

Mass of ice = 10 g

Temperature change= 10 °C

Specific heat = ?

Solution:

Formula:

Q = m. c. ΔT

Q = amount of heat

m = mass

c = specific heat

ΔT = change in temperature

Now we will put the values in formula,

Q = m. c. ΔT

200 j = 10 g. c . 10 °C

c = 200 j / 100 g. °C

c = 2 j/ g. °C

The specific heat of ice is 2 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.

58750.08068 is the answer i think