[mass Cu sample x specific heat Cu x (Tfinal-Tinitial)] + [mass liquid + specific heat liquid x (Tfinal-Tinitial)] + [mass container of Cu x specific heat Cu x (Tfinal-Tinitial)] = 0
Plus in the numbers and solve for specific heat liquid.
Temperature can be confusing so here is a summary of the data.
400 g Cu Tfinal-Tinitial. Start at 100 C and ends at 50 C.
10 g container of Cu. Tfinal-Tinitial. Start at 30C and ends at 50 C.
100 g liquid. Tfinal - Tinitial. Starts at 30 C and ends at 50 C.
A piece of copper weighing 400g is heated to 100°c and then quickly transferred to a copper calorimeter of Mass of 10g containing 100g of liquid of unknown specific heat capacity at 30°c if the final temperature of the mixture is 50°c calculate the specific heat capacity of liquid ( specific heat capacity if copper = 390j/kg/k)
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