Use Q = mcAT

1. How much heat (Q) will be absorbed by 320 g of water when its temperature is raised by

35°C?

The specific heat (C) for water is 4.18 J/gºC

2. Calculate the specific heat (c) for aluminum when 16500J of heat are absorbed in raising

the temperature of 150g of aluminum by 125°C

Question

Chemistry

Duncan Rivas

13 August, 21:04

Use Q = mcAT

1. How much heat (Q) will be absorbed by 320 g of water when its temperature is raised by

35°C?

The specific heat (C) for water is 4.18 J/gºC

2. Calculate the specific heat (c) for aluminum when 16500J of heat are absorbed in raising

the temperature of 150g of aluminum by 125°C

+2

Answers (1)

Know the Answer?

Kenna Williamson · Answer 1

1) ... for the specific heat capacity (c) of water is 4200kg/J°C ...

... guven mass (m) = 320g (0.32kg)

... change in temperature (ΔT) = 35°C

from the formula

Q=mcΔT

Q=0.32Kg x 4200kg/J°C x 35°C

Q=47,040Joules

Use Q = mcAT1. How much heat (Q) will be absorbed by 320 g of water when its temperature is raised by35°C?The specific heat (C) for water is 4.18 J/gºC2. Calculate the specific heat (c) for aluminum when 16500J of heat are absorbed in raisingthe temperature of 150g of aluminum by 125°C

Use Q = mcAT

1. How much heat (Q) will be absorbed by 320 g of water when its temperature is raised by

35°C?

The specific heat (C) for water is 4.18 J/gºC

2. Calculate the specific heat (c) for aluminum when 16500J of heat are absorbed in raising

the temperature of 150g of aluminum by 125°C