Question

The temperature of a sample of water increases from 30.0 degrees C to 40.0 degrees C as 6540 joules of heat are added. What is the mass of the sample of water?

Answer 1

Answer:

166 g

Explanation:

The equation of specific heat capacity is:

ΔE = mcΔtheta

ΔE = 6540J

c = 4200J/kg°C

Δtheta = 40 - 30 = 10

6540 = m × 4200 × 10

6540 = m × 42000

6540 ÷ 42000 = m

m = 0.15571428...

m = 0.166 kg

m = 166 g

Answer 2

Answer:  166 g

Explanation: q = mC∆T

q = heat = 6940 J

m = mass = ?

C = specific heat of water = 4.184 J/g/deg

∆T = change in temperature = 40.0 - 30.0 = 10.0ºC

6940 J = (m)(4.184J/g/deg)(10 deg)

m = 165.9 g = 166 g