Normal 0false false falseEN-US X-NONE X-NONE/* Style Definitions */table.MsoNormalTable{mso-style-name:Table Normal;mso-tstyle-rowband-size:0;mso-tstyle-colband-size:0;mso-style-noshow:yes;mso-style-priority:99;mso-style-parent:;mso-padding-alt:0in 5.4pt 0in 5.4pt;mso-para-margin-top:0in;mso-para-margin-right:0in;mso-para-margin-bottom:10.0pt;mso-para-margin-left:0in;line-height:115%;mso-pagination:widow-orphan;font-size:11.0pt;font-family:Calibri,sans-serif;mso-ascii-font-family:Calibri;mso-ascii-theme-font:minor-latin;mso-hansi-font-family:Calibri;mso-hansi-theme-font:minor-latin;mso-bidi-font-family:Times New Roman;mso-bidi-theme-font:minor-bidi;}10gm of ice cubes at 0°C are released in a tumbler (water equivalent 55 g) at 40°C. Assuming that negligible heat is taken from the surroundings, the temperature of water in the tumbler becomes nearly (L = 80 cal/g)(a) 31°C(b)22°C(c)19°C(d)15°C
Simran Bhatia , 11 Years ago
Grade 11
1 Answers
Amit Saxena
Last Activity: 11 Years ago
(b)
Let the final temperature be T
Heat gained by ice = mL + m × s × (T- 0)
= 10 × 80 + 10 × 1 × T
Heat lost by water = 55 × 1 × (40 - T)
By using law of calorimetery,
800 + 10T = 55 × (40-T)
Þ T = 21.54° C = 22°C
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