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

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