Peter pours 750g of water over ice in a cup. Determine the amount of heat lost b
ID: 1475446 • Letter: P
Question
Peter pours 750g of water over ice in a cup. Determine the amount of heat lost by the water of temperature 40°C as it melts 100g of ice of temperature -5°C (Lfusion= 333 J/g). At the time the ice has completely changed its state to liquid (but before the thermal equilibrium) what is the temperature of the water that was poured into the cup? What is the final temperature of the water poured and water from melted ice as they reach thermal equilibrium? The specific heat if water is 4.18 J/g°C. The specific heat of ice is 2.11J/g°C.
Explanation / Answer
the amount of heat lost by the water of temperature 40°C as it melts 100g of ice of temperature -5°C,
Q1 = m_ice*C_ice*(0 -(-5)) + m_ice*Lf
= 100*2.11*5 + 100*333
= 34355 J <<<<<<<<<------------Answer
Let T is the temperature of water at this time.
use, Q1 = m_water*C_water*(40 - T)
Q/(m_water*C_water) = 40 - T
T = 40 - Q/(m_water*C_water)
= 40 - 34355/(750*4.18)
= 29 degrees celsius <<<<<<<<<<<<<<-------------Answer
Let Tf is the final temperature of water poured.
use, amount heat gained by ice = amount of heat lost by water
m_ice*C_water*(Tf - 0) = m_water*C_water*(29 - Tf)
m_ice*Tf = m_water*29 - m_water*Tf
Tf*(m_ice + m_water) = m_water*29
Tf = m_water*29/(m_ice + m_water)
= 750*29/(100+750)
= 25.6 degrees celsius <<<<<<<<<<<<<<-------------Answer
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