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levacccp [35]
2 years ago
13

A (10.0+A) g ice cube at -15.0oC is placed in (125 B) g of water at 48.0oC. Find the final temperature of the system when equili

brium is reached.
Specific heat of ice: 2.090 J/g K
Specific heat of water: 4.186 J/gK
Latent heat of fusion for water: 333 J/g
Physics
1 answer:
liq [111]2 years ago
4 0

Answer: Final temperature is 34.15°C.

Explanation: When two objects have different temperature, they will exchange heat energy until there is no more net energy transfer between them. At that state, the objects are in <u>thermal</u> <u>equilibrium</u>.

So, when in equilibrium, the total heat flow must be zero, i.e.:

Q_{1}+Q_{2}=0

In our case, there will be a change in state of ice into water, so total heat flow will be:

m_{1}c_{1}(T_{f}-T_{i})+m_{2}c_{2}(T_{f}-T_{i})+mL=0

where

m₁ is mass of ice

m₂ is mass of water

c₁ is specific heat of ice

c₂ is specific heat of water

T_{f} is final temperature

T_{i} is initial temperature

L is latent heat fusion

Temperature is in Kelvin so the transformation from Celsius to Kelvin:

For ice:

T = -15 + 273 = 258K

For water:

T = 48 + 273 = 321K

Solving:

21(2.09)(T_{f}-258)+158(4.186)(T_{f}-321)+21(333)=0

43.89T_{f}-11323.62+661.4T_{f}-212305.55+6993=0

705.3T_{f}=216636.17

T_{f}= 307.15K

In Celsius:

T_{f}= 34.15°C

Final temperature of the system when in equilibrium is 34.15°C

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