Question

20 g ofice at -10°C is mixed with 10 g of water at o°C. The amount of heat required
to raise the temperature of mixture to 10°C is​

Answer 1

Answer:

Hi MATE..

271⁰C

hope it helps..❤✌..

Answer 2

$Q_{Total}=8343.8\ J$

When 20g of ice at -10°C is mixed with .......

Explanation:

Given that:

• mass of ice, $m_i=20\ g$

• mass of water, $m_w=10\ g$

• initial temperature of ice, $T_{ii}=-10^{\circ}C$

• initial temperature of water, $T_{iw}=0^{\circ}C$

• final temperature of mixture, $T_f=10^{\circ}C$

We have,

• latent heat of ice, $L=334\ J.g^{-1}$

• specific heat capacity of ice, $c_i=2.04\ J.g^{-1}.^{\circ}C^{-1}$

• specific heat capacity of water, $c_w=4.186\ J.g^{-1}.^{\circ}C^{-1}$

Now, assuming that there is no heat loss from the system of ice-water mixture:

Heat required for raising the temperature of the water.

$Q_w=m_w.c_w.\Delta T_w$

$Q_w=10\times 4.186\times 10$

$Q_w=418.6\ J$

Heat required for raising the temperature of the ice.

$Q_i=$ heat to bring to 0°C+heat of melting+heating of water from 0°C

$Q_i=20\times 2.04\times 10+20\times 334+20\times 4.186\times 10$

$Q_i=7925.2\ J$

Now, total heat required:

∴$Q_{Total}=Q_w+Q_i$

$Q_{Total}=418.6+7925.2$

$Q_{Total}=8343.8\ J$

#learnMore  TOPIC: Sensible Heat, Latent Heat

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