What will be the final temperature if 50g of water at exactly 0°C is added to 250g of water at 90°C?
Given:
"t_1=0^\\circ\\rm C"
"t_2=90^\\circ\\rm C"
"m_1=50\\:\\rm g"
"m_2=250\\:\\rm g"
The heat balance gives
Hence, the final temperature
"t_3=\\frac{m_1t_1+m_2t_2}{m_1+m_2}""t_3=\\frac{50*0+250*90}{50+250}=75^\\circ\\rm C"
Comments
Leave a comment