Answered: "An ice-cube tray of negligible mass contains 2.5..."

Answered question

2022-04-29

An ice-cube tray of negligible mass contains 2.5 kg of water at 20°C. How much heat must be removed to cool the water to 0°C and freeze it?

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Skilled2022-05-14Added 403 answers

$m_{w a t e r} = 2.5 k g$

$△ T = 0 - 20 = - 20 ° C$

$c_{w a t e r} = 4190 J / k g ° C$

$L_{f . w a t e r} = 3.34 * 10^{5} J / k g$

$Q_{g a i n} + Q_{l o s s} = 0$

$Q_{g a i n} + m_{w a t e r} * c_{w a t e r} * △ T - m_{w a t e r} * L_{f . w a t e r} = 0$

$Q_{g a i n} = - m_{w a t e r} * c_{w a t e r} * d e l t T + m_{w a t e r} * L_{f . w a t e r} = - 2.5 * 4190 * (- 20) * 2.5 * 3.34 * 10^{5}$

$= 1.749325 * 10^{11} J$

$= 165804018, 4 B t u$

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