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