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Answer 1 · 2017-03-21T16:20:55

The temperature was $=49.52ºC$

Let $T=$ final temperature of the water

For the water, $Delta T_w=50-T$

$m_w C_w (DeltaT_w)=2$

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

$1*4.186*(50-T)=2$

$50-T=2/4.186$

$50-T=0.48$

$T=50-0.48$

$T=49.52ºC$