The temperature of a sample of water increases by 69.5 degrees when

24,000 J of heat are applied. The specific heat of water is 4.18 J/gC. What

is the mass of the sample?

O

84.3

O

1692

407,356

7,117,495

The mass of the sample is 84.3 g

Explanation:

We have the formula to find the mass

Q = mCΔT

where

Q denotes the energy absorbed or released

m denotes the mass of the sample

C denotes the Specific heat of sample

Δ T denotes the Temperature change

Q = mCΔT

Rewrite the equation as,

m = Q / C Δ T

m=24000 J/4.186 J/g°C * (69.5 °C-0°C)

m=82.42

m=84.3 (based on the options)

The mass of the sample is 84.3 g