I would say B) hawk and coyote.
Hope this helps!
Answer:
0.886 J/g.°C
Explanation:
Step 1: Calculate the heat absorbed by the water
We will use the following expression
Q = c × m × ΔT
where,
- c: specific heat capacity
- ΔT: change in the temperature
Q(water) = c(water) × m(water) × ΔT(water)
Q(water) = 4.184 J/g.°C × 50.0 g × (34.4 °C - 25.36 °C) = 1.89 × 10³ J
According to the law of conservation of energy, the sum of the energy lost by the solid and the energy absorbed by the water is zero.
Q(water) + Q(solid) = 0
Q(solid) = -Q(water) = -1.89 × 10³ J
Step 2: Calculate the specific heat capacity of the solid
We will use the following expression.
Q(solid) = c(solid) × m(solid) × ΔT(solid)
c(solid) = Q(solid) / m(solid) × ΔT(solid)
c(solid) = (-1.89 × 10³ J) / 32.53 g × (34.4 °C - 100. °C) = 0.886 J/g.°C
The rate of evaporation would decrease because of
the formation of bonds between the soap and the water.
So, no. Regular water evaporates faster than soapy water
