The equilibrium temperature will be equal to 70^@"C"70∘C.
In order to solve this problem, you need to know the specific heat of water, of copper, and of glass.
c_"water" = "4.18 J/g"^@"C"cwater=4.18 J/g∘C
c_"copper" = "0.386 J/g"^@"C"ccopper=0.386 J/g∘C
c_"glass" = "0.84 J/g"^@"C"cglass=0.84 J/g∘C
So, the idea is that the heat lost by the copper will be absorbed by the water and the glass. The equilibrium temperature will thus be lower than 940^@"C"940∘C, and higher than 20^@"C"20∘C.
Mathematically, this is written as
q_"copper" = -(q_"water" + q_"glass")qcopper=−(qwater+qglass)
q_"copper" = -q_"water" - q_"glass"qcopper=−qwater−qglass color(blue)((1))(1)
The relationship between heat lost/gained and temperature change for a substance is given by
q = m * c * DeltaT, where
q - heat;
m - the mass of the substance;
c - its specific heat;
DeltaT - the change in temperature, defined as the difference between the equilibrium temperature and the initial temperature;
So, plug your data into equation color(blue)((1)) to get
m_"copper" * c_"copper" * (T_"ech" - T_"copper"^0) = -m_"water" * c_"water" * (T_"ech" - T_"water"^0) - m_"glass" * c_"glass" * (T_"ech" - T_"glass"^0)
1000cancel("g") * 0.386cancel("J")/(cancel("g") * ^@cancel("C")) * (T_"ech" - 940)^@cancel("C") = 1500cancel("g") * 4.18cancel("J")/(cancel("g") * ^@cancel("C")) * (T_"ech" - 20)^@cancel("C") - 500cancel("g") * 0.84cancel("J")/(cancel("g") ^@cancel("C")) * (T_"ech" - 20)^@cancel("C")
386 * T_"ech" - 362840 = -6270 * T_"ech" + 125400 - 720 * T_"ech" + 8400
7076 * T_"ech" = 496640 => T_"ech" = 496640/7076 = 70.19^@"C"
Rounded to one sig fig, the number of sig figs you gave for the mass of copper, the answer will be
T_"ech" = color(green)(70^@"C")
