Question

A(n) 70-g ice cube at 0°C is placed in 960 g of water at 24°C.
What is the final temperature of the mixture?

______ °C

Answer #1

we know that

heat lost by the water is equal to the heat gained by the ice from the principle of calorimetry

Q_{loss by water} = Q _{heat gained by ice}

m_{w}c_{w} ( 24^{o} - T ) =
m_{ice} L_{f} + m_{ice}c_{w} (T -
0^{o} C )

m_{w}c_{w} ( 24^{o} - T ) =
m_{ice} L_{f} + m_{ice}c_{w} T

m_{w}c_{w} ( 24^{o}) -
m_{w}c_{w} T = m_{ice} L_{f} +
m_{ice}c_{w} T

m_{ice}c_{w} T + m_{w}c_{w} T =
m_{ice} L_{f} - m_{w}c_{w} (
24^{o})

T = ( m_{w}c_{w} ( 24^{o})-
m_{ice} L_{f} ) / (
m_{ice}c_{w} + m_{w}c_{w} )

given data

m_{ice}=70 g = 70x10^{-3} kg

m_{water}= 960 g = 0.96 kg

c_{w}= 4.18x10^{3} J/kg^{o}c

L_{f}= 333.5x10^{3} J/ kg

plugging the values in the formula

T =
((0.96)*(4.18x10^{3})*(24)-(70x10^{-3})*(333.5x10^{3}))
/ ((0.96)*(4.18x10^{3}) +
(70x10^{-3})*(4.18x10^{3}) )

T =16.95^{o}C

**final temperature of the mixture = T =
16.95 ^{o}C**

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