210
Thermochemistry
raise 1 g of the liquid by 1°C (its specific heat) is
.. , t ... .,
155 cal
specific heat of liquid = (200.0 ml)(0.900 g/ml)(0.950°C)
= 0.910^
An alternative method of calorimetry that gives less accurate results, but is
simpler in concept, uses only a single insulated container and a thermometer.
Temperature changes in the calorimeter are brought about by adding hot (or
cold) objects of known weight and temperature. Calculations are based on the
principle that the heat lost by the added hot object is equal to that gained by the
water in the calorimeter and the calorimeter walls. This simple approach is
illustrated in the next two problems.
PROBLEM:
The temperature in a calorimeter containing 100 g of water is 22.7°C. Fifty grams
of water are heated to boiling (99.1°C at this location) and quickly poured into the
calorimeter. The final temperature is 44.8°C. From these data, calculate the heat
capacity of the calorimeter.
SOLUTION:
The heat loss from the hot water is equal to the heat gain by the calorimeter and
the water initially in it.
Heat lost by hot water = (wt of H 2 O)(sp ht of H 2 O)(temp change)
= (50 g) (l.OO ^} (99.1°C - 44.8°C)
= 2715 cal
Heat gained by calorimeter water = (wt of H 2 O)(sp ht of H 2 O)(temp change)
= (ioogm:oo -^M (44.8°c - 22.ro
= 2210 cal
Heat gained by calorimeter = (ht capacity of calorimeter)(temp change)
= (x ^rj (44.8°C - 22.TC)
= 22. IA cal
We equate the heat lost to the heat gained and solve for x, the heat capacity of the
calorimeter:
Thermochemistry
raise 1 g of the liquid by 1°C (its specific heat) is
.. , t ... .,
155 cal
specific heat of liquid = (200.0 ml)(0.900 g/ml)(0.950°C)
= 0.910^
An alternative method of calorimetry that gives less accurate results, but is
simpler in concept, uses only a single insulated container and a thermometer.
Temperature changes in the calorimeter are brought about by adding hot (or
cold) objects of known weight and temperature. Calculations are based on the
principle that the heat lost by the added hot object is equal to that gained by the
water in the calorimeter and the calorimeter walls. This simple approach is
illustrated in the next two problems.
PROBLEM:
The temperature in a calorimeter containing 100 g of water is 22.7°C. Fifty grams
of water are heated to boiling (99.1°C at this location) and quickly poured into the
calorimeter. The final temperature is 44.8°C. From these data, calculate the heat
capacity of the calorimeter.
SOLUTION:
The heat loss from the hot water is equal to the heat gain by the calorimeter and
the water initially in it.
Heat lost by hot water = (wt of H 2 O)(sp ht of H 2 O)(temp change)
= (50 g) (l.OO ^} (99.1°C - 44.8°C)
= 2715 cal
Heat gained by calorimeter water = (wt of H 2 O)(sp ht of H 2 O)(temp change)
= (ioogm:oo -^M (44.8°c - 22.ro
= 2210 cal
Heat gained by calorimeter = (ht capacity of calorimeter)(temp change)
= (x ^rj (44.8°C - 22.TC)
= 22. IA cal
We equate the heat lost to the heat gained and solve for x, the heat capacity of the
calorimeter: