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.

Question

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).

Answer 1

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.

Answer 2

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.

Answer 3

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