Answer:
The amount of time required to rise the temperature of 0.50 liter of water from 0.0 c to 10.0 c is 19 seconds
Explanation:
1100 J/s is the energy supplied so, we have to set, the energy to that ΔT, which is 10°C (10°-0°)
Q = m . C. ΔT
Q = 500g . 4,186J/g°C . 10°C
Remember the water density is 1g/mL, so if we have 0,5 L, we should get the convertion to g. (1ml = 1x10*-3 L)
1x10*-3 L ____ 1 g
0,5 L _____ (0,5 L . 1 g)/ 1x10*-3 L = 500g
4,186J/g°C, this value is known as specific heat of water. It is a known value.
Q = 500g . 4,186J/g°C . 10°C = 20930 J
Now the final rule of three:
1100 J supplies at _____ 1 second
20930 J supplies at _____ (20930 J . 1 s) / 1100 J =19,02 s
