A piece of glass has a temperature of 83.0 °C. Liquid that has a temperature of 43.0 °C is poured over the glass, completely covering it, and the temperature at equilibrium is 59.0 °C. The mass of the glass and the liquid is the same. Ignoring the container that holds the glass and liquid and assuming that the heat lost to or gained from the surroundings is negligible, determine the specific heat capacity of the liquid.

Not sure how to attempt this question but I know that I need to use the formula Q=CmdeltaT and that the specific heat capacity of the glass is = 840 J/kg(C).

Calculate the specific heat capacityof the materials for which the heat supply, mass and temperature rise 2.52×10^4j, 0.3kg, 20k

To solve this problem, you can use the principle of energy conservation.

First, let's identify the information given in the problem:

- Initial temperature of the glass = 83.0 °C
- Initial temperature of the liquid = 43.0 °C
- Temperature at equilibrium = 59.0 °C
- Specific heat capacity of the glass = 840 J/kg°C

We can apply the principle of energy conservation, which states that the heat gained by the liquid and the glass should be equal to the heat lost by the glass.

The heat gained by the liquid can be calculated using the formula Q = m * c * ΔT, where Q is the heat gained, m is the mass of the liquid, c is the specific heat capacity of the liquid, and ΔT is the change in temperature.

The heat lost by the glass can be calculated using the same formula, Q = m * c * ΔT, but now using the properties of the glass.

Since the mass of the glass and liquid is the same, m can be canceled out from both equations.

Now, we can set up the equation based on the principle of energy conservation:

m * c * ΔT (liquid) = m * c (glass) * ΔT (glass)

Using the given values, we have:

c (liquid) * ΔT (liquid) = c (glass) * ΔT (glass)

Now, we can plug in the values:

c (liquid) * (59.0 °C - 43.0 °C) = 840 J/kg°C * (83.0 °C - 59.0 °C)

Simplifying the equation:

c (liquid) * 16.0 °C = 840 J/kg°C * 24.0 °C

Dividing both sides by 16.0 °C:

c (liquid) = (840 J/kg°C * 24.0 °C) / 16.0 °C

c (liquid) = 1,260 J/kg°C

Therefore, the specific heat capacity of the liquid is 1,260 J/kg°C.

Not sure? one gains heat, one loses heat (a negative gain), so the sum of the heats gained is zero.

HeatGainedLiquid+HeatGainedGlassss=0
m(cliq)(43-59)+m(cglass)(83-59)=0
solve for c of liquid.

Since we ignore the mass, it becomes:

(cliq)*(-16) + (840)*(24) = 0

(cliq)*(-16) = -20160

(cliq) = 7.936*10^-4 ??
Answer is wrong, please check.