The equilibrium temperature will be equal to #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"#
#c_"copper" = "0.386 J/g"^@"C"#
#c_"glass" = "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"#, and higher than #20^@"C"#.
Mathematically, this is written as
#q_"copper" = -(q_"water" + q_"glass")#
#q_"copper" = -q_"water" - q_"glass"# #color(blue)((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")#
