<span>If Dr. Gavin decides that instead of conducting a 2 ´ 4 independent-groups factorial design, and he is going to conduct a 2 ´ 4 within-subjects factorial design, then the things that will change are the various independent groups which are involved.</span>
Answer:
5.01×10⁴ J.
Explanation:
Applying,
q = Cm....................... Equation 1
Where q = amount of heat needed to melt the ice, m = mass of the ice, C = specific latent heat of ice.
From the question,
Given: m = 150 g = (150/1000) kg = 0.15 kg, C = 3.34×10⁵ J/kg
Substitute these values into equation 1
q = (0.15×3.34×10⁵)
q = 0.501×10⁵ J
q = 5.01×10⁴ J.
<span>Assuming pulley is frictionless. Let the tension be ‘T’. See equation below.</span>
<span> </span>
Answer:
= 85.7 ° C
Explanation:
For this exercise we will use the calorimetry heat ratios, let's start with the heat lost by the evaporation of coffee, since it changes from liquid to vapor state
Q₁ = m L
Where m is the evaporated mass (m = 2.00 103-3kg) and L is 2.26 106 J / kg, where we use the latent heat of the water
Q₁ = 2.00 10⁻³ 2.26 10⁶
Q1 = 4.52 10³ J
Now the heat of coffee in the cup, which does not change state is
Q coffee = M
(
-
)
Since the only form of energy transfer is terminated, the heat transferred is equal to the evaporated heat
Qc = - Q₁
M ce (
-
) = - Q₁
The coffee dough left in the cup after evaporation is
M = 250 -2 = 248 g = 0.248 kg
-Ti = -Q1 / M
= Ti - Q1 / M 
Since coffee is essentially water, let's use the specific heat of water,
= 4186 J / kg ºC
Let's calculate
= 90.0 - 4.52 103 / (0.248 4.186 103)
= 90- 4.35
= 85.65 ° C
= 85.7 ° C