Question

1. What was the mass of a sample of water if the addition of 4280 joules raised the temperature of the sample from 22.00C to 34.50C?

2. What is the specific heat of a sample of an alloy if 3750 joules of heat are released when a 315 gram sample of alloy cools from 78.00C to 28.40C?
3. The specific heat of iron is 0.448 J/g0C. What will be the final temperature if 35000 joules of heat are added to a 454 gram sample of iron at 24.00C?
4. What will be the final temperature when 8750 joules are added to 75.0 grams of water at 23.00C?
5. When 38800 joules were added to a sample of iron, specific heat 0.448 J/g0C, the temperature rose from 24.5 deg C to 178 deg C. What was the mass of the iron sample?

Answer 1

Step-by-step explanation:

Number 1

Q = 4280J

mass = x

∆∅ = 34.5 - 22

∆∅ = 12.5°C

Specific heat capacity of water = 4200J

Q = mc∆∅

4280 = x(4200)(12.5)

4280 = 52500x

x = 4280/52500

x = 0.0815kg

Number 2

c = x

Q = 3750

m = 315g

∆∅ = 78 - 28.4

∆∅ = 49.6°C

Q = mc∆∅

3750 = 315×x×49.6

3750 = 15624x

x = 3750/15624

x = 0.2400J(g°C)-¹

Number 3

c = 0.448

Q = 35000

m = 454g

∆∅ = x

Q = mc∆∅

35000 = 454(0.448)(x)

35000 = 203.392x

x = 35000/203.392

x = 172.0815

∅2 - ∅1 = ∆∅

∅2 = ∆∅ + ∅1

∅2 = 172.0815 + 24

∅2 = 196.0815

∅2 = 196°C

Number 4

Q = 8750J

m = 75g = 0.075

c = 4200J/kg/°C

∆∅ = x

Q = mc∆∅

8750 = 0.075(4200)∆∅

8750 = 315×x

x = 8750/315

x = 27.7778

∅2 - ∅1 = 27.7778

∅2 = 27.7778 + ∅1

∅2 = 27.7778 + 23

∅2 = 50.7778

∅2 = 50.8°C

Number 5

Q = 38800J

c = 0.448J/(g°C)

∆∅ = 178 - 24.5

∆∅ = 153.5°C

m = x

Q = mc∆∅

38800 = x × 0.448 × 153.5

38800 = 68.768x

x = 38800/68.768

x = 564.2159

x = 564g