Answer:
The heat absorbed by water is 39243.75 J.
Explanation:
Given data:
Mass of water = 375.0 g
Heat absorbed by water= ?
Initial temperature = 10.0°C
Final temperature = 35.0 °C
The specific heat capacity of water = 4.186 j/g.°C
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
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
ΔT = T2 - T1
ΔT = 35°C - 10°C
ΔT = 25°C
Q = m.c. ΔT
Q = 375.0 g× 4.186 j/g °C × 25°C
Q = 39243.75 J
The heat absorbed by water is 39243.75 J.
