A student adds 46.7 g of water to a calorimeter and allows the temperature to stabilize in a thermostat bath at 25.00°C.
Next, s/he adds 43.8 g of 100.0°C water.
The final temperature of the entire apparatus comes to 49.3°C.
What is the effective heat capacity of the calorimeter in J/°C?
Take the specific heat capacity of water as 4.184 J g-1°C-1
Hint: heat is lost by the hot water; heat is gained by both the cold water and the calorimeter.
Express the answer in the form ±x.xxE±x
Heat Capacity of Calorimeter? PLZZZZZZZ Help!?
heat lost by hot water = m*c*t ( mass* sp heat cap* temp diff) = 43.8* 4.184*(100- 49.3) J
heat gained by water= 46.7* 4.184* (49.3-25) J
heat gained by calorimetre = mass? * c * (49.3-25)
mass of calorimetre is not given in the problem without which the prob cant be solved.
heat lost by hot water = heat gained by cold water plus calorimetre
sp heat cap of calorimetre = (heat lost by hot water - heat lost by cold water)/ mass of calorimetre * temp diff of calorimetre
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