A piece of copper ball of mass 20g at 200°is place in a copper calorimeter of 60g containing 50g of water at 39°.ignoring heat calculate the final steady temperature of mixture (specific heat capacity water=4400J/Kg/°k
You did not say what the heat capacity of copper is.
I am not about to look it up.
Call it Kc Joules / GRAM deg C (convert to grams if your table is for Kg)
You have 20 g of Cu at 200
You have 60 g of Cu at 39
You have 50 g of H2O at 39 Kw = 4.400 J/ GRAM deg K or C
final temp = T
heat out = Kc (20) (200-T)
heat in = 4.4 * 50 (T-39) + Kc *60 *(T-39)
heat out = heat in, solve for T
To calculate the final steady temperature of the mixture, we need to assume that there is no heat lost to the surroundings (i.e., the system is isolated) and that the heat gained by the cooler objects is equal to the heat lost by the hotter objects.
First, let's calculate the heat lost by the copper ball:
Heat lost by the copper ball = mass of the copper ball × specific heat capacity of copper × change in temperature
= 20 g × specific heat capacity of copper × (final temperature - initial temperature)
= 20 g × 387 J/kg °C × (final temperature - 200°C)
Next, let's calculate the heat gained by the water and the calorimeter:
Heat gained by the water and the calorimeter = (mass of water + mass of calorimeter) × specific heat capacity of water × change in temperature
= (50 g + 60 g) × 4400 J/kg °C × (final temperature - 39°C)
Since we assume that the heat gained is equal to the heat lost, we can set up an equation:
20 g × 387 J/kg °C × (final temperature - 200°C) = (50 g + 60 g) × 4400 J/kg °C × (final temperature - 39°C)
Now, we can solve this equation to find the final steady temperature of the mixture.