If an 83.00 g sample of iron has a starting temperature of 297k and an ending temperature of 329k how much heat will be lost from the iron samples? if the heat absorbed from a 59.00 g sample is 4814.4 j and the temperature change is 85.0c, what is the samples specific heat? what is the sample? the heat absorbed by a sample of 55.00 g is 3908.3 j and its starting temperature is 78.0c and its ending temperature is 95.0c, what is the specific heat of this sample and what is the sample?

Question 1 :

m = mass of iron sample = 83

c = specific heat of iron sample = 0.450 J/g ⁰C

T₀ = initial temperature = 297 K

T = final temperature = 329 K

Q = heat lost

Heat lost is given as

Q = m c (T - T₀)

inserting the values

Q = (83) (0.450) (329 - 297) = 1195.2 J

Question 2 :

m = mass of sample = 59 g

c = specific heat of sample = ?

ΔT = change in temperature = 85 c

c = specific heat

Q = heat absorbed = 4814.4 J

Heat absorbed is given as

Q = m c

inserting the values

4814.4 = (59) (85) c

c = 0.96 J/g C

the sample is cholorofom

Question 3 :

m = mass of sample = 55 g

c = specific heat of sample = ?

T₀ = initial temperature = 78 c

T = final temperature = 95 c

Q = heat absorbed = 3908.3 J

Heat absorbed is given as

Q = m c (T - T₀)

inserting the values

3908.3 = (55) (95 - 78) c

c = 4.2 J/g C

the sample is water

the force required to move the cart is divided by three or 33.33 n.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s