Respuesta :
Answer:
1.50 cal
Explanation:
Given data:
Mass of water = 300 g
Initial temperature = 20 °C
Final temperature = 25 °C
Calories needed = ?
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Solution:
ΔT = T2 - T1
ΔT = 25 °C - 20°C
ΔT =
Q = m.c. ΔT
Q = 300 g . 4.18j/g. °C . 5°C
Q = 6270 J
J calories:
6270 / 4184 = 1.50 cal