Question

When 519 J up he is added to a sample of liquid water the temperature rises from 21.6 Celsius to 61.8 Celsius given the specific heat capacity of liquid water is 4.184J/G•K how many grams of water are in the sample

Answer 1

ΔT = 61.8 – 21.6 = 40.2 ºC

Q = mcΔT

$m = \frac{Q}{cΔT} \\ = \frac{519}{4.184 \times 40.2} \\ = \frac{519}{168.19} \\ = 3.08 \;g$

Answer: 3.08 g