Answer:
a) T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c) 
b) T = 295.37 K
Explanation:
Given;
Initial temperature of tea T1 = 31 C
Initial temperature of ice T2 = 0 C
Mass of tea m1 = 0.89 kg
Mass of ice m2 = 0.075kg
The heat capacity of both water and tea c = 4186 J/(kg⋅K)
the latent heat of fusion for water is Lf = 33.5 × 10^4 J/kg
And T = the final temperature of the mixture
Heat loss by tea = heat gained by ice
m1c∆T1 = m2c∆T2 + m2Lf
m1c(T1-T) = m2c(T-T2) + m2Lf
m1cT1 - m1cT = m2cT - m2cT2 + m2Lf
m1cT + m2cT = m1cT1 + m2cT2 - m2Lf
T(m1c + m2c) = m1cT1 + m2cT2 - m2Lf
T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c) 
Substituting the values;
T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)
T = (0.89×4186×31 + 0.075×4186×0 - 0.075×33.5 × 10^4)/(0.89×4186 + 0.075×4186)
T = 22.37 °C
T = 273 + 22.37 K
T = 295.37 K
 
        
                    
             
        
        
        
0 (zero) work done.
This is because work can be calculated by:
Work = Force × Distance
The force is 100 N and since the rock didn't move the distance is zero and so:
Work = 100 × 0 = 0 J
 
        
             
        
        
        
Andaniol I’m for sure of it
        
             
        
        
        
Answer:
c 
Explanation:
I makes the most sense in math. if you put a 3 in the m box it will be the same. So all they did is take out the 3 in the m box