Answer:
The required specific heat is 196.94 joule per kg per °C
Step-by-step explanation:
Given as :
The heat generated = Q = 85.87 J
Mass of substance (m)= 34.8 gram = 0.0348 kg
Change in temperature = T2 - T1 = 34.29°C - 21.76°C = 12.53°C
Let the specific heat = S
Now we know that
Heat = Mass × specific heat × change in temperature
Or, Q = msΔt
Or, 85.87 = (0.0348 kg ) × S × 12.53°C
Or , 85.87 = 0.4360 × S
Or, S =
∴ S = 196.94 joule per kg per °C
Hence the required specific heat is 196.94 joule per kg per °C Answer
Answer:
Just factorize the given plynomials.
Step-by-step explanation:
like in first example 7x+49
7(x+7)
Answer:
I don’t know the area but here’s a hint to help you (hint: length x with x height)
Step-by-step explanation:
Answer:
28
Step-by-step explanation: