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
9514 1404 393
Answer:
$8,775
Step-by-step explanation:
The amount due is given by the formula ...
A = P(1 +rt)
where P is the principal amount, r is the annual rate, and t is the number of years.
A = $6,500(1 +0.07×5) = $6,500(1.35) = $8,775
Montrey had to pay back $8,775.
Make the same denominator (12) and then make them both non mixed numbers 93/12 and 54.6666.../12
so 147.666.../12
12.305/12
Answer:
6 kw are used each hour
Step-by-step explanation:
E = P t
E/t = P
66 kwh /11h = 6 kw