Try this solution:
if given m=0.15 kg; t₁=20 °C; t₂=100 °C; c=4190 J/(kg*C); q=226*10⁴ J/kg., then
Q=Q₁+Q₂,
where Q₁=cm(t₂-t₁) and Q₂=q*m.
Finally,
Q=cm(t₂-t₁)+qm;
Q=4190*0.15*80+2240000*0.15=386280 J=<u>386.28 kJ</u>.
Violet i took the test and it gave me the answer
Answer:
M2 = 278.06 kg
Explanation:
We calculate the weight of M1
W=m*g
Where
m: mass (kg)
g: acceleration due to gravity (m/s²)
W₁=288* 9.8= 2822.4 N
Look at the attached graphic
We calculate the x-y components of the weight :
W₁x= 2822.4*sin41° N =1851.66 N
W₁y= 2822.4 *cos41° N = 2130.09 N
We apply Newton's first law for the balance in M1:
Σ Fy=0
Fn-W₁y=0 , Fn: normal force
Fn=W₁y=2130.09N
Friction Force = Ff=μs *Fn = 0.41*2130.09 =873.34 N
Σ Fx=0
T- W₁x- Ff=0
T= 1851.66 + 873.34
T= 1851.66 + 873.34
T=2725 N
We apply Newton's first law for the balance in M2:
Σ Fy=0
T- W₂ =0
W₂ = T = 2725 N
W₂ = M2*g
M2 = W₂/g
M2 = 2725/9.8
M2 = 278.06 kg
Heat always flows from higher temperature to lower temperature. So option A. heat will flow from the air into the coolant is the correct answer.