50 g of liquid X at 10 Celcius and 200 g of liquid Y
mx*cx*(t-tx)+my*cy*(t-ty)=0
cx/cy = - my*(t-ty) : mx*(t-tx) = (my/mx) * (ty - t) / (t-tx)
cx/cy = 200/50*(40-15)/(15-10) = 20
cx/cy = 20
Answer:
The extension of the wire is 0.362 mm.
Explanation:
Given;
mass of the object, m = 4.0 kg
length of the aluminum wire, L = 2.0 m
diameter of the wire, d = 2.0 mm
radius of the wire, r = d/2 = 1.0 mm = 0.001 m
The area of the wire is given by;
A = πr²
A = π(0.001)² = 3.142 x 10⁻⁶ m²
The downward force of the object on the wire is given by;
F = mg
F = 4 x 9.8 = 39.2 N
The Young's modulus of aluminum is given by;

Where;
Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²

Therefore, the extension of the wire is 0.362 mm.
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Q= mcΔT
Where Q is heat or energy
M is mass, c is heat capacitance and t is temperature
You have to convert Celsius into kelvin in order to use this formula I believe
Celsius + 273 = Kelvin
21 + 273 = 294K
363 + 273 = 636K
Now...
Q= (0.003)(0.129)(636-294)
Q= 0.132 J if you are using kilograms, in terms of grams which seems more appropriate the answer would be 132J of energy.