An object with a mass of $6 kg$ , temperature of $315 ^oC$ , and a specific heat of $12 (KJ)/(kg*K)$ is dropped into a container with $39 L$ of water at $0^oC$ . Does the water evaporate? If not, by how much does the water's temperature change?

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Answer 1 · 2017-07-25T09:12:56

The water does not evaporate and the change in temperature is $=96.4^@C$

The heat is transferred from the hot object to the cold water.

Let $T=$ final temperature of the object and the water

For the cold water, $Delta T_w=T-0=T$

For the object $DeltaT_o=315-T$

$m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)$

$C_w=4.186kJkg^-1K^-1$

$C_o=12kJkg^-1K^-1$

$6*12*(315-T)=39*4.186*T$

$315-T=(39*4.186)/(6*12)*T$

$315-T=2.267T$

$3.267T=315$

$T=315/3.267=96.4^@C$

As $T<100^@C$ , the water does not evaporate

An object with a mass of 6 kg6 kg, temperature of 315 ^oC315 ^oC, and a specific heat of 12 (KJ)/(kg*K)12 (KJ)/(kg*K) is dropped into a container with 39 L 39 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?