Q1) What heat flow in Joules is necessary to change the temperature of 0.50 kg o
ID: 1487481 • Letter: Q
Question
Q1) What heat flow in Joules is necessary to change the temperature of 0.50 kg of water from
25 °C to 45 °C? [Hint: the specific heat capacity of water is 4186 J kg C ]
Q2) In the previous problem, a heat flow changed the temperature of 0.50 kg of water from 25 °C to 45 °C. If the specific heat capacity of water was larger than its typical value of
4186 J kg C , which of the following statements would be true?
(A) The change in temperature of the water would be greater. (B) Less energy is needed to change the temperature from 25 °C to 45 °C. (C) The water’s temperature would not change (D) More energy is needed to change the temperature from 25 °C to 45 °C.
Q3) A heat flow of 45, 000 J changed the temperature of a certain mass of water from 25 °C to 45 °C. If the heat flow came from 350 g of lead shot, what must have been the heat flow, Q, in
Joules for the lead shot? [Hint: the specific heat capacity of water is 4186 J kg C , and the
specific heat capacity of lead is 128 J kg C ]
Q4) Heated lead shot is added to cold water, and the temperature of the water rises. If the heat flow into the water, Q, is 3850 J and the mass of the lead shot is 350 g, what is the temperature
change of the lead shot (in °C)? [Hint: the specific heat capacity of water is 4186 J kg C , and
the specific heat capacity of lead is 128 J kg C ]
Q5) When two thermodynamic systems at different temperatures are brought into thermal contact and are otherwise isolated from their surroundings, the sum of the heat flows for the two systems is (A) negative. (B) equal to zero. (C) positive. (D) undefined.
Explanation / Answer
Ans 1
heat flow
Q = mC(T2-T1)
Q = 0.50*4186*(45-25)
Q = 41860 J
Ans 2
If specific heat of water is greater then its typical value then more heat is required to change the temperature.
D) More energy is needed to change the temperature from 25 °C to 45 °C.
Ans 3
Here heat for lead must be equal to the heat flow for water so
Q = 45000 J
ans 4
let T2-T1 is temperature change of lead shot then
3850 = mC(T2-T1)
T2 - T1 = 3850/0.350*128
T2 - T1 = 85.93 degree celcius
ans 5
When two thermodynamic systems at different temperatures are brought into thermal contact and are otherwise isolated from their surroundings, the sum of the heat flows for the two systems is equal to zero. as heat absorbed by one system = heat supplied by other
(B) equal to zero
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